The below are some of example queries to which users can find how many ways to arrange letters in a word by using this word permutation or letters arrangement calculator:
- how many ways to arrange 2 letters word?
- how many different ways to arrange 3 letters word?
- how many distinguishable ways to arrange 4 letters word?
- how many ways are there to order the letters LAKES?
- how many distinguishable permutations of the letters of the word STATISTICS?
- how many different ways can the letters of the word MATHEMATICS be arranged?
- in how many ways can the letters of the word MATH be arranged?
- find the number of distinct permutations of the word LAKES?
- how many ways can the letters in the word LOVE be arranged?
Just give a try the words such as HI, FOX, ICE, LOVE, KIND, PEACE, KISS, MISS, JOY, LAUGH, LAKES, MATH, STATISTICS, MATHEMATICS, COEFFICIENT, PHONE, COMPUTER, CORPORATION, YELLOW, READ and WRITE to know how many ways are there to order the 2, 3, 4, 5, 6, 7, 8, 9 or 10 letters word. Users also supply any single word such as name of country, place, person, animal, bird, ocean, river, celebrity, scientist etc. to check how many ways the alphabets of a given word can be arranged by using this letters arrangement or permutation calculator.
Below is the reference table to know how many different ways to arrange 2, 3, 4, 5, 6, 7, 8, 9 or 10 letters word can be arranged, where the order of arrangement is important. The n-factorial (n!) is the total number of possible ways to arrange a n-distinct letters word or words having n-letters with some repeated letters. Refer permutation formula to know how to find nPr for different scenarios such as:
- finding word permutation for words having distinct letters,
- finding word permutation for words having repeated letters.
| Number of Ways to Arrance ‘n’ Letters of a Word | |
|---|---|
| ‘n’ Letters Words | Ways to Arrange |
| 2 Letters Word | 2 distinct ways |
| 3 Letters Word | 6 distinct ways |
| 4 Letters Word | 24 distinct ways |
| 5 Letters Word | 120 distinct ways |
| 6 Letters Word | 720 distinct ways |
| 7 Letters Word | 5,040 distinct ways |
| 8 Letters Word | 40,320 distinct ways |
| 9 Letters Word | 362,880 distinct ways |
| 10 Letters Word | 3,628,800 distinct ways |
Work with Steps: How many Distinct Ways to Arrange the Letters of given Word
Supply the word of your preference and hit on FIND button provides the answer along with the complete work with steps to show what are all the parameters and how such parameters and values are being used in the permutation formula to find how many ways are there to order the letters in a given word. Click on the below words and know how the calculation is getting changed based on the word having distinct letters and words having repeated letters. For other words, use this letters of word permutations calculator.
- FLORIDA
- GEORGIA
- CALIFORNIA
- NEVADA
- MARYLAND
- MONTANA
Enter Word
How does the Letter Arrangements in a Word Calculator work?
Given a word, this determines the number of unique arrangements of letters in the word.
This calculator has 1 input.
What 1 formula is used for the Letter Arrangements in a Word Calculator?
- Arrangements = M!/N1!N2!…NM!
For more math formulas, check out our Formula Dossier
What 3 concepts are covered in the Letter Arrangements in a Word Calculator?
- factorial
- The product of an integer and all the integers below it
- letter arrangements in a word
- permutation
- a way in which a set or number of things can be ordered or arranged.
nPr = n!/(n — r)!
Letter Arrangements in a Word Calculator Video
First select the places which the vowels are going to occupy. There are only three vowels so there are $3!=6$ ways to choose these (we need to assign to each column the row of that column that is going to have the vowel, these three numbers must be distinct, so $3!$)
Once we do that there are $3$ ways to decide how to place the vowels inside the selected places (Since this is equivalent to selecting the position for the letter O, and we have three options).
This tells us there are $3cdot 6=18$ ways to place the vowels.
Once the vowels have been placed we have to place the consonants, however we can do this however we want. There are $6$ places remaining and $6$ consonants, so naively we could say there are $6!$ ways to place the consonants. However in reality we have one letter $P$, two letters $R$ one letter $F$ one letter $M$ and one letter $D$.
This means if we consider the $6!$ arrangements we are counting each arrangement twice, since switch the places of the letter $R$ with the other letter $R$ gives us the same arrangement. Hence there are $frac{6!}{2}=360$ ways to place the consonants (See multinomial coefficient for more information on this).
All in all there $360cdot18=6480$ ways to place the letters.
Permutations/Arrangements
The number of permutations with n different things taking r at a time is given by nPr, where
| nPr | = | |
| = | (n)! × (n − 1)! × (n − 2)! × …. r times |
Permutations/Arrangements in making words from letters
nL : Number of letters with which the words are to be formed
nP : Number of places to fill the letters i.e. the number of letters in the word to be formed
nP ≤ nL
Since the maximum length of the word that can be formed is limited to the number of letters available
Number of nP lettered words that can be formed using nL letters where all the letters are different
⇒ Number of words that can be formed using nL letters taking nP letters at a time
⇒ Number of permutations of nL items taking nP at a time
| ⇒ nLPnP | = | |
| = | (nL) × (nL − 1) × (nL − 2) × …. nP times |
Explanation
Forming a nP letter word with nL letters can be assumed as the act of arranging nL letters into nP places.
| 1st | 2nd | … | … | nPth |
- Total Event (E)
Arranging the nL letters in nP places
The total event can be divided into the sub-events of placing each letter in a place starting from the first.
- 1st sub-event (SE1)
Placing a letter in the First place
The 1st place can be filled with any one of the available nL letters
⇒ SE1 can be accomplished in nL ways
⇒ nSE1 = nL
- 2nd sub-event (SE2)
Placing a letter in the Second place
The 2nd place can be filled with any one of the remaining nL − 1 letters
⇒ SE2 can be accomplished in nL − 1 ways
⇒ nSE2 = nL − 1
- 3rd sub-event (SE3)
Placing a letter in the Third place
The 3rd place can be filled with any one of the remaining nL − 2 letters
⇒ SE3 can be accomplished in nL − 2 ways
⇒ nSE3 = nL − 2
- …
- …
- nPth sub-event (SEnP)
Placing a letter in the nPth place
The nPth place can be filled with any one of the nL − (nP − 1) letters remaining after filling the first nP − 1 places
⇒ SE3 can be accomplished in nL − (nP − 1) ways
⇒ nSEnp = nL − nP + 1
By the fundamental counting principle of multiplication,
Number of ways in which the nL letters can be filled in the nP places
| ⇒ nE | = | nSE1 × nSE2 × nSE3 × …. × nSEp |
| = | (nL) × (nL − 1) × (nL − 2) × … × (nL − nP + 1)
= (nL) × (nL − 1) × (nL − 2) × … nP times ≡ n × (n − 1) × (n − 2) × … × (n − r + 1) |
|
| = | nLPnP
≡ nPr |
Example
The number of 5 letter words that can be formed with the letters of the word subdermatoglyphic
In the word subdermatoglyphic
Number of letters
= 17
{S, U, B, D, E, R, M, A, T, O, G, L, Y, P, H, I, C}
⇒ nL = 17
Number of letters in the word to be formed
⇒ Number of places to be filled in forming the word
= 5
⇒ nP = 5
Number of words that can be formed with the letters of the word subdermatoglyphic
= Number of words that can be formed using nL letters taking nP letters at a time
= Number of permutations of nL items taking nP at a time
= nLPnP
= 17P5
= 17 × 16 × 15 × 14 × 13
Where all letters are used
Where all letters are used in forming the words,
nL = nP
Number of ways in which the nL letters can be filled in the nP places
| ⇒ nE | = | nLPnP |
| = | nLPnL | |
| = | nL! | |
| Or | = | nPPnP |
| = | nP! |
Example
The number of words that can be formed with the letters of the word Algebra
In the word Algebra
Number of letters
= 7
{A, L, G, E, B, R, A}
⇒ nL = 7
Number of letters in the word to be formed
⇒ Number of places to be filled in forming the word
= 7
⇒ nP = 7
Number of words that can be formed with the letters of the word Algebra
= Number of words that can be formed using nL letters taking nP letters at a time
= Number of permutations of nL items taking nP at a time
= nLPnP
= nL!
= 7!
= 4,090
Fixing letters (each in its own place)
nL : Number of letters with which the words are to be formed
nP : Number of places to fill the letters i.e. the number of letters in the word to be formed
nP ≤ nL
Since the maximum length of the word that can be formed is limited to the number of letters available
nFL : Number of letters to be fixed each in its own place
After fixing nFL letters each in its own place
Number of Letters remaining
= Total number of letters − Number of letters to be fixed
⇒ nRL = nL − nFL
Number of Places remaining to be filled
= Total number of places − Number of letters to be fixed
⇒ nRP = nP − nFL
Number of nP letter words that can be formed using nL letters by fixing nFL letters each in its own place is given by
nRLPnRP
Explanation
- Total Event (E)
Arranging the nL letters in nP spaces
Assume that the total event is divided into two sub-events.
- 1st sub-event (SE1)
Arranging the nFL letters each in its own place
Number of ways in which this event can be accomplished
= 1
Since each letter can be placed in a specific place only there is only one way this can be done whatever may be the number of letters being fixed
⇒ nSE1 = 1
- 2nd sub-event (SE2)
Filling the remaining nRP places with the remaining nRL letters
Number of ways in which this event can be accomplished
= Number of ways in which nRP places can be filled with the nRL letters Or = Number of permutations or arrangements of nRL items taking nRP items at a time
The number of ways in which the nL letters can be filled in the nP places with nFL letters fixed each in its own place
| ⇒ nE | = | nSE1 × nSE2 |
| = | 1 × nRLPnRP | |
| = | nRLPnRP |
Fundamental Counting principle of Multiplication
If a total event can be sub divided into two or more sub events all of which are independent, then the total number of ways in which the total event can be accomplished is given by the product of the number of ways in which each sub event can be accomplished.
Example
Number of 7 letter words that can be formed with the letters of the word incomputably such that T occupies the middle place
In the word incomputably
Number of letters in the word
= 12
{I, N, C, O, M, P, U, T, A, B, L, Y}
nL = 12
Number of places to be filled in forming the words
= 7
nP = 7
Number of letters fixed each in its own place
nFL = 1
After fixing the specified letters in their respective places
Number of letters remaining
= Total Number of letters − Number of letters fixed in specific places
| ⇒ nRL | = | nL − nFL |
| = | 12 − 1 | |
| = | 11 |
Number of places remaining to be filled
= Total Number of Places − Number of letters fixed in specific places
| ⇒ nRP | = | nP − nFL |
| = | 7 − 1 | |
| = | 6 |
The number of 7 letter words that can be formed with the letters of the word incomputably such that T occupies the middle place
= Number of ways in which 12 letters can be arranged in 7 places such that 1 letter is fixed in a specified place
= nRLPnRP
= 11P6
= 11 × 10 × 9 × 8 × 7 × 6
using fundamental principle
- Total Event (E)
Filling the 7 places with the 12 letters
- 1st sub-event (SE1)
Filling the middle place with T
Number of ways in which this event can be accomplished
= Number of ways in which the 1 place can be filled with the 1 letter Or = Number of permutations or arrangements with 1 item taking 1 at a time - 2nd sub-event (SE2)
Filling the remaining places with the remaining letters
Number of ways in which this event can be accomplished
= Number of ways in which the remaining 6 places can be filled with the remaining 11 letters Or = Number of permutations or arrangements with 11 items taking 6 at a time
The number of 7 letter words that can be formed with the letters of the word incomputably such that T occupies the middle place
| ⇒ nE | = | nSE1 × nSE2 |
| = | 1 × 11P6 | |
| = | 11P6 | |
| = | 11 × 10 × 9 × 8 × 7 × 6 |
Where all letters are used
Where all letters are used in forming the words,
nL = nP
⇒ nL − nFL = nP − nFL
⇒ nRL = nRP
Number of nP lettered words that can be formed with the nL letters, fixing nFP letters each in its own place
| = | nRLPnRP | |
| = | nRLPnRL | |
| = | nRL! | |
| Or | = | nRPPnRP |
| = | nRP! |
Example
The number of words that can be formed with the letters of the word Thursday such that T occupies the first place and Y occupies the last place
In the word thursday,
Number of letters
= 8
{T, H, U, R, S, D, A, Y}
nL = 8
Number of places to be filled in forming the words
= 8
nP = 8
Number of letters fixed each in its own place
nFL = 2
After fixing the specified letters in their respective places
Number of letters remaining
= Total Number of letters − Number of letters fixed in specific places
| ⇒ nRL | = | nL − nFL |
| = | 8 − 2 | |
| = | 6 |
Number of places remaining to be filled
= Total Number of Places − Number of letters fixed in specific places
| ⇒ nRP | = | nP − nFL |
| = | 8 − 2 | |
| = | 6 |
Number of words that can be formed with the letters of the word Thursday such that T occupies the first place and Y occupies the last place
= Number of ways in which 8 letters can be arranged in 8 places such that 2 letters are fixed each in its specified place
= nRLPnRP
= nRLPnRL Or nRPPnRP
= nRL! or nRP!
= 6!
= 720
using fundamental principle
- Total Event (E)
Filling the 8 places with the 8 letters
- 1st sub-event (SE1)
Filling the first place with T and last place with Y
Number of ways in which this event can be accomplished
= Number of ways in which the 2 places can be filled with the 2 specified letters each in its own place
= 1
⇒ nSE1 = 1
- 2nd sub-event (SE2)
Filling the remaining places with the remaining letters
Number of ways in which this event can be accomplished
= Number of ways in which nRP remaining places can be filled with the nRL remaining letters Or = Number of permutations or arrangements with nRP items taking all at a time ⇒ nSE2 = nRLPnRP = 6P6 = 6! = 720
The number of words that can be formed with the letters of the word thursday that start with T and end with Y
| ⇒ nE | = | nSE1 × nSE2 |
| = | 1 × 720 | |
| = | 720 |
Fixing a set of (two or more letters) in a set of places
nL : Number of letters with which the words are to be formed
nP : Number of places to fill the letters i.e. the number of letters in the word to be formed
nP ≤ nL
The maximum length of the word that can be formed is limited to the number of letters available.
nSL : Number of letters specified to be filled in the specified places
nSP : Number of places specified to be filled with the specified letters
nFP : Number of places filled
If nSL ≠ nSP,
Number of places filled would be the lesser of nSL and nSP.
If the specified letters are lesser only that many letters can be used up for filling and if the specified places are lesser only that many places can be filled.
⇒ nFP = smaller of nSL and nSP
Where nSL = nSP,
nFP = nSP = nSL
After fixing nFL letters each in its own place
Number of Letters remaining
= Total number of letters − Number of letters to be fixed
⇒ nRL = nL − nFL
Number of Places remaining to be filled
= Total number of places − Number of letters to be fixed
⇒ nRP = nP − nFL
nRP ≤ nRL
The places remaining to be filled cannot exceed the letters remaining to be used
Number of nP letter words that can be formed using nL letters by fixing nFL letters each in its own place is given by
nRLPnRP
Explanation
- Total Event (E)
Arranging the nL letters in nL spaces
Assume that the total event is divided into two sub-events.
- 1st sub-event (SE1)
Arranging the nSL specified letters in the nSP specified places
Number of ways in which this event can be accomplished
= Number of ways in which the nSP specified places can be filled with the nSL specified letters Or = Number of permutations or arrangements with greater of nSP and nSL items taking the lesser of them at a time ⇒ nSE1 = aPb Number of permutations of a items taking b at a time
Where
a = larger of nSL and nSP and b the other
If nSL = nSP,
a = b = nSL = nSP
- 2nd sub-event (SE2)
Arranging the remaining letters in the remaining places.
Number of ways in which this event can be accomplished
= Number of ways in which nRP places can be filled with the nRL letters Or = Number of permutations or arrangements of nRL items taking nRP at a time
The number of words that can be formed
| ⇒ nE | = | nSE1 × nSE2 |
| = | aPb × nRLPnRP
Where a = larger of nSL and nSP and b the other |
Examples
The number of words that can be formed with the letters of the word Equation such that the even spaces are occupied by vowels
In the word Equation
Number of letters
= 8 {E, Q, U, A, T, I, O, N}
⇒ nL = 8
In the word to be formed
Number of places
= 8
⇒ nP = 8
Number of specified letters
⇒ Number of Vowels
= 5
{E, U, A, I, O}
⇒ nSL = 5
Number of specified places
⇒ Number of Even places
= 4
{X, _, X, _, X, _, X, _}
⇒ nSP = 4
After filling the even places with vowels
Number of places filled
= 4
Smaller of nSL (5) and nSP (4)
⇒ nFP = 4
Number of Places remaining to be filled
= Total Places − Places Filled
| ⇒ nRP | = | nP − nFP |
| = | 8 − 4 | |
| = | 4 |
Number of Letters remaining to be used
= Total Letters − Places Filled
| ⇒ nRL | = | nL − nFP |
| = | 8 − 4 | |
| = | 4 |
Number of words that can be formed with the letters of the word Equation such that the even spaces are occupied by vowels
| = | aPb × nRLPnRP
Where a is the larger of nSL (5) and nSP (4) and b the other |
| = | 5P4 × 4P4 |
| = | (5 × 4 × … 4 terms) × 4! |
| = | (5 × 4 × 3 × 2) × (4 × 3 × 2 × 1) |
| = | 120 × 24 |
| = | 2,880 |
Using counting principles
- Total Event (E)
Filling the 8 places with the 8 letters
- 1st sub-event (SE1)
Filling the 4 even places with the 5 letters
Number of ways in which this event can be accomplished
= Number of ways in which 4 places can be filled with 5 letters Or = Number of permutations or arrangements with 5 letters taking 4 at a time ⇒ nSE1 = 5P4 = 5 × 4 × 3 × 2 = 120 - 2nd sub-event (SE2)
Filling the remaining places with the remaining letters
Number of ways in which this event can be accomplished
= Number of ways in which the remaining 4 places can be filled with the remaining 4 letters Or = Number of permutations or arrangements with 4 letters taking 4 at a time
Number of words that can be formed with the letters of the word equation such that even places are occupied by vowels
| ⇒ nE | = | nSE1 × nSE2 |
| = | 120 × 24 | |
| = | 2,880 |
The number of 8 letter words that can be formed with the letters of the word warehousing such that odd positions have only consonants.
In the word hypnotizable
Number of letters
= 11
{W, A, R, E, H, O, U, S, I, N, G}
⇒ nL = 11
In the words to be formed
Number of places
Number of letters
= 8
⇒ nP = 8
Number of specified letters
⇒ Number of Consonants
= 6
{W, R, H, S, N, G}
⇒ nSL = 6
Number of specified places
⇒ Number of Odd Positions
= 4
{X, _, X, _, X, _, X, _}
⇒ nSP = 4
After filling the odd places with consonants
Number of places filled
= 4
Smaller of nSL (6) and nSP (4)
⇒ nFP = 4
Number of Places remaining to be filled
= Total Places − Places Filled
| ⇒ nRP | = | nP − nFP |
| = | 8 − 4 | |
| = | 4 |
Number of Letters remaining to be used
= Total Letters − Places Filled
| ⇒ nRL | = | nL − nFP |
| = | 11 − 4 | |
| = | 7 |
Number of 8 letter words that can be formed with the letters of the word warehousing such that the odd positions are filled with consonants
| = | aPb × nRLPnRP
Where a is the larger of nSL (6) and nSP (4) and b the other |
| = | 6P4 × 7P4 |
| = | (6 × 5 × … 4 terms) × (7 × 6 × … 4 terms) |
| = | (6 × 5 × 4 × 3) × (7 × 6 × 5 × 4) |
| = | 360 × 840 |
Using counting principles
- Total Event (E)
Filling the 8 places with the 11 letters
- 1st sub-event (SE1)
Filling the 6 consonants in the 4 odd positions
Number of ways in which this can be accomplished
= Number of ways in which 4 positions can be filled with 6 letters Or = Number of permutations or arrangements of 6 items taking 4 at a time ⇒ nSE1 = 6P4 = 6 × 5 × 4 × 3 = 360 - 2nd sub-event (SE2)
Filling the remaining 4 places with the remaining 7 letters
Number of ways in which this event can be accomplished
= Number of ways in which 4 places can be filled with the 7 letters Or = Number of permutations or arrangements with 7 items taking 4 at a time ⇒ nSE2 = 7P4 = 7 × 6 × 5 × 4 = 840
The number of words that can be formed with the letters of the word warehousing such that odd positions are filled with consonants
| ⇒ nE | = | nSE1 × nSE2 |
| = | 360 × 840 |
Two or more letters grouped (stay together)
nL : Number of letters with which the words are to be formed
nP : Number of places to fill the letters i.e. the number of letters in the word to be formed
nP ≤ nL
The maximum length of the word that can be formed is limited to the number of letters available.
nGL1 : Number of letters in the first group
nGL2 : Number of letters in the second group
nG : Number of groups of letters
| nGL | : | Number of letters considering the letters to be grouped as a unit |
| = | Total number of letters − Number of letters grouped + Number of groups of letters
Remove all the letters that are grouped and add a letter for each group. |
| ⇒ nGL | = | (nL − ∑i=1nGnGLi) + nG |
Number of nP lettered words that can be formed using nL letters such that nGL1 stay as a group, nGL2 stay as another group, … .
- Total Event (E)
Arranging the nL letters in nP spaces
Assume that the total event is divided into nG + 1 sub-events.
- 1st sub-event (SE1)
Arranging the nGL letters in as many places,
Number of ways in which this event can be accomplished
= Number of ways in which nGL places can be filled with as many letters Or = Number of permutations or arrangements with nGL items taking all at a time - 2nd sub-event (SE2)
Arranging the letters in the first group among themselves.
Number of ways in which this event can be accomplished
= Number of ways in which nGL1 places can be filled with as many letters Or = Number of permutations or arrangements with nGL1 items taking all at a time ⇒ nSE2 = nGL1PnGL1 = nGL1! - 3rd sub-event (SE3)
Arranging the letters in the second group among themselves.
Number of ways in which this event can be accomplished
= Number of ways in which nGL2 places can be filled with as many letters Or = Number of permutations or arrangements with nGL2 items taking all at a time ⇒ nSE3 = nGL2PnGL2 = nGL2! - …
- …
The number of words that can be formed
| ⇒ nE | = | nSE1 × nSE2 × nSE3× … |
| = | nGL! × nGL1! × nGL2! × … |
Examples
The number of words that can be formed with the letters of the word Victory such that all the vowels come together
In the word Victory
Number of letters
= 7 {V, I, C, T, O, R, Y}
⇒ n = 7
Number of Vowels
= 2 {I, O}
⇒ Number of letters in the group = 2
⇒ g = 2
Number of letters considering the vowels as a unit
= 6 {V, (I,O), C, T, R, Y}
| ⇒ nG | = | 6 |
| Or | = | (n − g) + 1 |
| = | (7 − 2) + 1 | |
| = | 6 |
Number of words that can be formed with the letters of the word Victory such that all the vowels come together
| = | nG! × g! |
| = | 6! × 2! |
| = | 1,440 |
Using counting principles
- Total Event (E)
Filling the 7 places with the 7 letters
- 1st sub-event (SE1)
Arranging the letters taking the vowels as a unit in as many places
Number of ways in which this event can be accomplished
= Number of ways in which 6 places can be filled with as many letters Or = Number of permutations or arrangements with 6 letters taking all at a time - 2nd sub-event (SE2)
Inter arranging the two vowels among themselves
Number of ways in which this event can be accomplished
= Number of ways in which 2 places containing vowels can be filled with 2 vowels Or = Number of permutations or arrangements with 2 letters taking all at a time
Number of words that can be formed with the letters of the word victory such that all the vowels are together
| ⇒ nE | = | nSE1 × nSE2 |
| = | 720 × 2 | |
| = | 1,440 |
The number of words that can be formed with the letters of the word Daughters such that all the vowels come together and DGHT stay together.
In the word Daughter
Number of letters
= 9 {D, A, U, G, H, T, E, R, S}
⇒ n = 9
Number of Vowels (first group)
= 3 {A, U, E}
⇒ g1 = 3
Number of letters in the second group
= 4 {D, G, H, T}
⇒ g2 = 4
Number of groups
⇒ N = 2
Number of letters considering the each of the letters grouped as a unit
= 4 {(A,U,E), (D,G,H,T), R, S}
| ⇒ nG | = | 4 |
| Or | = | [n − (g1 + g2)] + N |
| = | [9 − (3 + 4)] + 2 | |
| = | 9 − 7 + 2 | |
| = | 4 |
Number of words that can be formed with the letters of the word Daughters such that all the vowels come together and DGHT stay together
| = | nG! × g1! × g2! |
| = | 4! × 3! × 4! |
| = | 24 × 6 × 24 |
| = | 3,456 |
Using counting principles
- Total Event (E)
Filling the 9 places with the 9 letters
- 1st sub-event (SE1)
Arranging the letters taking the vowels as a unit and DGHT as a unit in as many places
Number of ways in which this event can be accomplished
= Number of ways in which 4 places can be filled with as many letters Or = Number of permutations or arrangements with 4 letters taking all at a time - 2nd sub-event (SE2)
Inter arranging the vowels among themselves
Number of ways in which this event can be accomplished
= Number of ways in which 3 places can be filled with as many letters Or = Number of permutations or arrangements with 3 letters taking all at a time - 3rd sub-event (SE3)
Inter arranging the three letters DGHT among themselves
Number of ways in which this event can be accomplished
= Number of ways in which 4 places can be filled with as many letters Or = Number of permutations or arrangements with 4 letters taking all at a time
The number of words that can be formed with the letters of the word daughters such that all the vowels are together and the letters DGHT are together
| ⇒ nE | = | nSE1 × nSE2 × nSE3 |
| = | 24 × 6 × 24 | |
| = | 3,456 |
No two letters to come together
nL : Number of letters with which the words are to be formed
nP : Number of places to fill the letters i.e. the number of letters in the word to be formed
nP ≤ nL
The maximum length of the word that can be formed is limited to the number of letters available.
nDL : Number of letters to stay divided/separate
| nOL | : | Number of other letters |
| = | Total number of letters − Number of letters to stay divided/separate | |
| = | (nL − nDL) |
Places to arrange letters to stay separate
Using
- one letter we can keep two letters separate.
- two letters we can keep three letters separate
OL OL - …
- n letters we can keep n + 1 letters separate
Thus in finding the number of places available to place the letters to stay separate, consider an arrangement of other letters with places on either side
| OL1 | OL2 | OL3 | OL4 |
The places on either side of the other letters are the places where the letters to be divided/separated can appear to ensure that they do not come together.
| nDP | : | Number of places available to place the letters to say divided/separate |
| = | Number of other letters + 1 | |
| = | nOL + 1 |
nDP ≥ nDL
To be able to keep nDL letters separate we need at least as many spaces (nDP) to fill them up.
If nDP < nDL, then it would not be possible to ensure that the nDL letters stay separate.
-
nDP < nDL
Eg : Arrange the letters of the word UTOPIA such that no two vowels come together.
In the word UTOPIA
Number of letters
= 6
{U, T, O, P, I, A}
⇒ nL = 6
Number of Letters to stay separate
⇒ Number of Vowels
= 4
{U, O, I, A}
⇒ nDL = 4
Number of other letters
= Total number of letters − Number of letters to stay separate
⇒ nOL = nL − nDL = 6 − 4 = 2 Number of places to place the letters to stay separate
= Number of other letters + 1
⇒ nDP = nOL + 1 = 4 + 1 = 5 Since nDP < nDL, it would not be possible to arrange the letters in such a way that the vowels do not come together.
U T O P I A There are only two other letters, T and P. They can separate a maximum of 3 vowels. The fourth vowel would have to come beside another vowel.
-
nDP = nDL
When nDP = nDL, the specified letters would stay separate only if the word starts as well as ends with one of the letters to stay separate.
Eg : Arrange the letters of the word Anxious such that no two vowels come together.
In the word Fortune
Number of letters
= 7
{A, N, X, I, O, U, S}
⇒ nL = 7
Number of Letters to stay separate
⇒ Number of Vowels
= 4
{A, I, O, U}
⇒ nDL = 4
Number of other letters
= Total number of letters − Number of letters to stay separate
⇒ nOL = nL − nDL = 7 − 4 = 3 Number of places to place the letters to stay separate
= Number of other letters + 1
⇒ nDP = nOL + 1 = 3 + 1 = 4 Since nDP ≥ nDL, it would be possible to arrange the letters in such a way that the vowels do not come together.
A N I X O S U We can neither start nor end the word with one of the other letters.
-
nOL = nDL or nDP = nDL + 1
When nOL = nDL, ensuring that the letters to stay separate are arranged in the places specified for them would result in the other letters also staying separate.
Eg : Arrange the letters of the word NATURE such that no two vowels come together.
In the word NATURE
Number of letters
= 6
{N, A, T, U, R, E}
⇒ nL = 6
Number of Letters to stay separate
⇒ Number of Vowels
= 3
{A, U, E}
⇒ nDL = 3
Number of other letters
= Total number of letters − Number of letters to stay separate
⇒ nOL = nL − nDL = 6 − 3 = 3 Number of places to place the letters to stay separate
= Number of other letters + 1
⇒ nDP = nOL + 1 = 2 + 1 = 3 Since nDP ≥ nDL, it would be possible to arrange the letters in such a way that the vowels do not come together.
A N U R E T N U R E T A The letters would stay separated whether the word starts with a letter from the letters to stay separate or other letters.
- Total Event (E)
Arranging the nL letters in nP spaces
Assume that the total event is divided into two sub-events.
- 1st sub-event (SE1)
Arranging the nOL letters in nOL places.
Number of ways in which this event can be accomplished
= Number of ways in which nOL places can be filled with as many letters Or = Number of permutations or arrangements with nOL items taking all at a time - 2nd sub-event (SE2)
Arranging the nDL letters in the nDP places that would ensure their staying separate,
Number of ways in which this event can be accomplished
= Number of ways in which nDP places can be filled with nDL letters Or = Number of permutations or arrangements with nDP items taking nDL at a time
The number of words that can be formed
| ⇒ nE | = | nSE1 × nSE2 |
| = | nOL! × nDPPnDL |
Examples
The number of words that can be formed with the letters of the word DIALOGUE such that no two consonants come together
In the word DIALOGUE
Number of letters
= 8
{D, I, A, L, O, G, U, E}
⇒ nL = 8
Number of Letters to stay separate
⇒ Number of Consonants
= 3
{D, L, G}
⇒ nDL = 3
Number of other letters
= Total number of letters − Number of letters to stay separate
| ⇒ nOL | = | nL − nDL |
| = | 8 − 3 | |
| = | 5 |
Number of places to place the letters to stay separate
= Number of other letters + 1
| ⇒ nDP | = | nOL + 1 |
| = | 5 + 1 | |
| = | 6 |
Since nDP ≥ nDL, it would be possible to arrange the letters in such a way that the vowels do not come together.
Number of words that can be formed with the letters of the word DIALOGUE such that no two consonants come together
= nOL! × nDPPnDL
= 5! × (5 + 1)P3
= (5 × 4 × 3 × 2 × 1) × 6P3
= 120 × (6 × 5 × … 3 terms)
= 120 × (6 × 5 × 4)
= 120 × 120
= 14,400
Using Counting Principles
- Total Event (E)
Arranging the 8 letters in 8 spaces
- 1st sub-event (SE1)
Arranging the 5 other letters in 5 places.
Number of ways in which this event can be accomplished
= Number of ways in which 5 places can be filled with as many letters Or = Number of permutations or arrangements with 5 items taking all at a time ⇒ nSE1 = 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120 - 2nd sub-event (SE2)
Arranging the 3 letters in the 5 + 1 places around the other letters that would ensure their staying separate,
Number of ways in which this event can be accomplished
= Number of ways in which 6 places can be filled with 3 letters Or = Number of permutations or arrangements with 6 items taking 3 at a time ⇒ nSE2 = 6P3 = 6 × 5 × … 3 times = 6 × 5 × 4 = 120
The number of words that can be formed
| ⇒ nE | = | nSE1 × nSE2 |
| = | 120 × 120 | |
| = | 14,400 |
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Precalculus Mathematics Homework Help
Arrangement of letters in a word
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Thread starter
rajeshmarndi -
Start date
Jul 9, 2015
- Jul 9, 2015
- #1
Homework Statement
In how many ways 4-lettered words that can be formed using the letters of the word «BOOKLET».
The book solution solve it this way.
The word contains 7 letters out of which there are 2 O’s. So there are 6 letters.
∴ The number of 4 lettered words
= 7P4 — 6P4 = 840 — 360 = 480
Homework Equations
The Attempt at a Solution
What we need is to deduct the duplicate counting of 2 O’s from the total permutation.
So how does 6P4 solve this purpose.
Instead of 6P2,
isn’t it should have been
( 4P2 * 5P2 ) / 2
i.e number of ways these 2 O’s can be placed in 4 place * number of ways B,K,L,E,T can be placed in the remaining 2 places divide by 2.
= (12 * 20)/2 =120
And the answer would be 840 — 120 = 720 and not 480.
Answers and Replies
- Jul 9, 2015
- #2
- Jul 9, 2015
- #3
Instead of 6P2
isn’t it should have been
Correction it is 6P4
It is not only the words with two Os that are double counted in 7P4.
But I do not see any other letters being doubled.
My question, is the book solution, correct, any way?
- Jul 9, 2015
- #4
But I do not see any other letters being doubled.
Imagine the two Os in different colours. In all the words corresponding to 7P4, which words only differ in colour pattern?
(I agree with the book answer, though I’ve yet to understand how they obtain it by subtracting 6P4.)
- Jul 10, 2015
- #5
Imagine the two Os in different colours.
I understood that and that’s why I mentioned them as O1 & O2.
In all the words corresponding to 7P4, which words only differ in colour pattern?
I am not sure if I’ve understood you correctly, if we imagine the two Os in different colours(and each unique letter to be a different colour), then 7P4 will result all to be in different colour.[/QUOTE]
(I agree with the book answer, though I’ve yet to understand how they obtain it by subtracting 6P4.)
Where am I wrong in my below solution.
Instead of 6P2,
isn’t it should have been
( 4P2 * 5P2 ) / 2
i.e number of ways these 2 O’s can be placed in 4 place * number of ways B,K,L,E,T can be placed in the remaining 2 places divide by 2.= (12 * 20)/2 =120
And the answer would be 840 — 120 = 720 and not 480.
- Jul 10, 2015
- #6
Where am I wrong in my below solution.
Consider BLO1K, BLO2K.
- Jul 10, 2015
- #7
Imagine the two Os in different colours. In all the words corresponding to 7P4, which words only differ in colour pattern?
(I agree with the book answer, though I’ve yet to understand how they obtain it by subtracting 6P4.)
I called them O and O’ in my mid, the principle is the same. I too have not understood how they arrive at it being 6P4 other than numerical coincidence. I solved it by first considering all words as double counted and then adding those that this removes by error (i.e., half of the 4 letter words made from BKLET).
- Jul 10, 2015
- #8
Yes, thanks, it just didn’t hit my mind.
- Jul 10, 2015
- #9
As 7P4 — 6P4 = number of words containing 2 Os (including O1 & O2)
, since 6P4 = total permutation without 2 Os in the word.
- Jul 10, 2015
- #10
But, I am still not yet clear.As 7P4 — 6P4 = number of words containing 2 Os (including O1 & O2)
, since 6P4 = total permutation without 2 Os in the word.
Yes, but it’s coincidence. 360 have no Os, 240 have one O, 180 have two. It just happens that 360=(240+180)/2.
I can make an argument for subtracting 46P3. Consider the number that have a specific O (whether or not they have both).
- Jul 10, 2015
- #11
360 have no Os, 240 have one O, 180 have two.
5! = 120 has no Os, 4*5*4*3 = 240 has one O, (4!/(2!)^2)*5*4 = 120 has two Os. The sum of all of these is 480, alternatively you subtract 240+120 = 360 from 840 and get the same result.
Average of 240 and 180 is 360?
- Jul 10, 2015
- #12
5! = 120 has no Os, 4*5*4*3 = 240 has one O, (4!/(2!)^2)*5*4 = 120 has two Os. The sum of all of these is 480, alternatively you subtract 240+120 = 360 from 840 and get the same result.Average of 240 and 180 is 360?
I shouldn’t post while watching Le Tour.
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Letter Arrangements in Words
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13 Jun 2022, 10:45
[Aptitude] PnC: How many Ways to arrange letters of word «RECUPERATE» (Permutation without Formulas)
When you form words using alphabets (or letters), the ‘order’ matters.
Three alphabets T,A,C
Now you form words: CAT is not same as ACT; both words have different meanings hence order matters. This is a permuation problem.
- Case 1: All letters are different
- Case 2: Word contains repeating alphabets
- Case 3: More than one letter reappears in Word
- Case 4: Multipicking Consonents and Vowels
- Readymade Formulas for Word-arrangement problems
- Recommended Booklist for Aptitude in CSAT,CMAT,CAT,IBPS
- Previous articles on Aptitude
Case 1: All letters are different
- Question: How many words can you form with 3 letters: A, C, T? OR
- How many ways can the letters of word “ACT” be arranged? OR
- Rephrase: How many ways can three gentlemen Abdul, Champaklal and Tarak Mehta (A, C, T) be seated in three chairs?
If you want to visualize, OR Manually arrange them (Desi Jugaad) it’ll look like this:
Otherwise it is a mere extention of Fundamental counting principle
3 candidates for first seat AND then 2 guys remain for second seat AND only one guy remains for the last seat
“AND” means multiplication
=3 x 2 x 1 ; meaning 3! ways
= 6
Readmady Formula: When “n” different items or alphabets or men are to be arranged in a line, it can be done in n! factorial ways.
“ACT” has three alphabets so it can be done in three factorial (3!=3x2x1 ways)
This was easy because all three alphabets A,C,T were different so you can apply the readymade formula.
But what if there is repeatation? For example the word “Repeat” itself!
Case 2: Word contains repeating alphabets
Q. How many ways can the letters of word “REPEAT” be arranged?
To prevent mistakes in such question, Better make a frequency chart as shown below
R…I
E…II
P…I
A…I
T…I
The letter “E” appears twice, hence although we 6 letters word, there are only 5 “different” letters or alphabets.
lets label these two “E”s as E1 and E2.
Form a word
REEPAT
It can be formed in two ways
RE1E2AT
or
RE2E1AT
Although both have same meaning.
When we arrange letters and form words, according to Funda.counting or Permutation method, these two words will be counting as ‘two different’ words. We’ve to remove this overcounting.
So we divide the answer with overcounting. Just like how we proceeded in Combination question in the very first article of PnC.
Step:1 Assume all alphabets to be different
We’ve 6 alphabets
Arrange these 6 gentlemen into 6 seats? (Order matters, Permutation problem)
=6 x 5 x 4 x 3 x 2 x 1 =6P6=6! ways.
Out of these two gentlemen are same. How many ways can your arrange two men in two seats?
=2 options for the first seat AND then 1 guy remains for the second seat
=2 x 1 ; AND means multiplication
=2! ways
Final answer
=6! divided by overcounting
=6!/2!
=360.
Let us now make the situation even more complex: What if multiple alphabets of a word, are getting repeated?
Case 3: More than one letter reappears in Word
Q. How many ways can the letters of word “RECUPERATE” be arranged? -(From Sarvesh Kumar’s book)
Make frequency character to prevent mistakes
R…II
E…III
C…I
U…I
P…I
A…I
E…I
Total 10 letters but R appears twice (II) and E appears thrice (III)
Principle is same like previous case
First make permutation of all letters, assuming that they’re different. And then divide by overcounting.
So Break it in tasks
- Task 1: Arrange 10 letters (10!)
- Task 2: Divide by overcounting of Two Rs (2!)
- Task 3: AND Again divide by overcounting of Three Es (3!)
- Task 4: keep repeating…if more letters are repeated
Start Solving
=10!/2! ;Task 1
=10!/2! ; Task 2 again divide by 3!
=10!/(2! x 3!) ; Task 3
=302400 ways letters of the word “RECUPERATE” can be arranged.
CASE 4: Multipicking Consonents and Vowels
Question from Indiabix on request of a reader.
Q. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Photo for better visualization of this concept
Words of 3 consonants and 2 vowels= 3+2=5 letters.We’ve to pick up some members in the Committee and then arrange them in ‘seats’.
Break it like this
Task 1 AND Task 2 AND Task 3
(Pick 3 conso out of given 7) AND (pick 2 vowels out of given 4) AND (Arrange them in 5 seats)
=Combi AND Combi AND Permu
=Combi x Combi x Permu ; because “AND” means multiplication
==7C3 x 4C2 x 5!
Task 1: Pick 3 conso out of given 7
Same as Committee problem.
Pick three men out of 7:
7 x 6 x 5
but in Committee, order doesn’t matter so Divide the overcounting
(arranging the selected 3 men in three seats)
3 x 2 x 1
So 7C3= (7 x 6 x 5) / (3 x 2 x 1)
Task 2: Pick 2 Vowels out of given 4
Same way you do for 4C2
Thus we selected 3 + 2 = 5 letters. (alphabets)
Task 3: Arrange 5 letters in a word
Suppose you’ve three letters A, C and T. You start forming words-
CAT= is not same as ACT.
Both words have different meanings so order matters, this is a permuation problem.
So How many ways can you form words using 5 letters or alphabets?
=How many ways can you make 5 gentlemen sit in 5 chairs (permutation problem, as seen in first article on PnC)
5 x 4 x 3 x 2 x 1 =5! ways.
Now gather everything together in one place
Task 1 AND Task 2 AND Task 3
7C3 x 4C2 x 5!
=25200
Readymade Formulas for Word-arrangement problems
- When all letters of the word are different: Number of permutation =n! ; where “n” is the number of letters.
But why does it work? Because we are doing permutation of “n” Men in “n” chairs and nPn=n!
- When the word contains “n” letters, out of which P1 are alike and are of one type, P2 are alike and of second type and P3 are alike and of third type and all the rest are different, then number of permutations
=n!/ (P1! x P2! x P3!)
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| 01. |
The letters of the word TRIANGLE are arranged at random. Find the probability and odds that the word so formed (i) starts with T (ii) ends with R (iii) starts with T and ends with R. Solution » Hide/Show Number of Letters/Characters in the word probablity = 8 {T, R, I, A, N, G, L, E} In the experiment of testing for the number of words that can be formed using the letters of the word «TRIANGLE» Total Number of Possible Choices = Number of words that can be formed using the 8 letters of the word «TRIANGLE»
Let «A», «B», «C» be the events of the word so formed starts with «T», ends with «R», and starts with «T» and ends with «R» respectively For Event «A» No. of Favourable/Favorable Choices = The no. of words that can be formed using the letters of the word «TRIANGLE» Hide/Show Explanation In the word «Triangle» No. of letters = 8 {T, R, I, A, N, G, L, E} ⇒ No. of places = 8
∴ The number of words that can be formed with the letters of the word triangle that start with «T»
⇒ mA = (n − 1)! ⇒ mA = (8 − 1)! ⇒ mA = 7! ⇒ mA = 5,040 Probability that the word formed using all the letters of the word «TRIANGLE» starts with «T»
Number of UnFavorable Choices = Total Number of possible choices − Number of Favorable choices
⇒ mAc = n − mA ⇒ mAc = 35,280
OddsOdds in Favour of the words formed with the letters of the word «TRIANGLE» starting with «T»
Odds against the words formed with the letters of the word «TRIANGLE» starting with «T»
For Event «B» Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word «TRIANGLE» Hide/Show Explanation In the word «Triangle» No. of letters = 8 {T, R, I, A, N, G, L, E} ⇒ No. of places = 8
∴ The number of words that can be formed with the letters of the word triangle that end with «T»
⇒ mB = (n − 1)! ⇒ mB = (8 − 1)! ⇒ mB = 7! ⇒ mB = 5,040 Probability that the word formed using all the letters of the word «TRIANGLE» ends with «R»
For Event «C» Number of Favourable/Favorable Choices = The no. of words that can be formed using the letters of the word «TRIANGLE» by fixing Hide/Show Explanation In the word «Triangle» No. of letters = 8 {T, R, I, A, N, G, L, E} ⇒ No. of places = 8
∴ The number of words that can be formed with the letters of the word triangle that end with «T»
⇒ mC = (n − 2)! ⇒ mC = (8 − 2)! ⇒ mC = 6! ⇒ mC = 720 Probability that the word formed using all the letters of the word «TRIANGLE» starts with «T» and ends with «R»
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| 02. |
The letters of the word �ORIENTAL� are arranged in all posible ways. Find the probability and odds of getting an arrangement which starts with an vowel and ends with a consonant. Solution » Hide/Show Number of Letters/Characters in the word probablity = 8 {O, R, I, E, N, T, A, L} In the experiment of testing for the number of words that can be formed using the letters of the word «ORIENTAL» Total Number of Possible Choices = Number of words that can be formed using the 8 letters of the word «ORIENTAL»
Let «A» be the event of the word so formed starting with a vowel and ending with a consonant For Event «A» No. of vowels = 4 {O, I, E, A} Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word «ORIENTAL» ⇒ mA = (No. of ways in which the first place can be filled with the 4 vowels) remaining 6 letters)
Probability that the word formed using all the letters of the word «ORIENTAL» starts with a vowel and ends with a consonant
Number of UnFavorable Choices = Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA ⇒ mAc = 40,320 − 11,520 (Or) 8! − (16 × 6!)
OddsOdds in Favour of the words formed starting with a vowel and ending with a consonant ⇒ Odds in Favor of Event «A» = Number of Favourable Choices : Number of UnFavorable Choices = 2 : 5 Odds against the words formed starting with a vowel and ending with a consonant = Number of Unfavorable Choices : Number of Unfavourable Choices = 5 : 2 |
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| 03. |
In a random arrangement of the letters of the word �uncopyrightable�. Find the probability and odds that all the vowels come together. Find also the probability that the vowels do not come together Solution » Hide/Show Number of Letters/Characters in the word «uncopyrightable» = 15 {U, N, C, O, P, Y, R, I, G, H, T, A, B, L, E} In the experiment of testing for the number of words that can be formed using the letters of the word «uncopyrightable» Total Number of Possible Choices = Number of words that can be formed using the 15 letters of the word «uncopyrightable»
Let «A» be the event of all the vowels coming together and «B» the event of the vowels not coming together For Event «A» No. of vowels (a) = 5 {U, O, I, A, E} Considering all the vowels as a unit No. of letters = 11 {N, C, (U, O, I, A, E), P, Y, R, G, H, T, B, L} Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word ⇒ mA = (No. of ways in which the 11 places can be filled with the letters (considering between themselves)
Probability that the word formed using all the letters of the word «uncopyrightable» have all the vowels together
Number of UnFavorable Choices = Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA ⇒ mAc = 15! − (11! × 5!)
OddsOdds in Favour of the words formed having all the vowels together = Number of Favourable Choices : Number of UnFavorable Choices = 11! × 5! : 11! × 32,640 Odds against the words formed having all the vowels together = Number of Unfavorable Choices : Number of Unfavourable Choices = mAc : mA For Event «B» Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word = Total number of words that can be formed ⇒ mB = n − mA ⇒ mB = mAc ⇒ mB = 11! × 32,640 Probability that the word formed not having all the vowels together
Alternatively Probability that the word formed not having all the vowels together ⇒ Probability of occurance of Event «B» = Probability of non-occurance of Event «A»
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| 04. |
The letters of SUCCESS are arranged at random. The probability that the vowels occupy even places is Solution » Hide/Show Number of Letters/Characters in the word «SUCCESS» = 7 {S, U, C, C, E, S, S} ⇒ nL = 7 No. of Letters : of the first kind ⇒ No. of S’s = 3 ⇒ a = 3 of the second kind ⇒ No. of C’s = 2 ⇒ b = 2 In the experiment of testing for the number of words that can be formed using the letters of the word «SUCCESS» Total Number of Possible Choices = Number of words that can be formed using the 8 letters of the word «SUCCESS»
Let «A» be the event of the vowels occupying even places For Event «A» No. of vowels = 2 {U, E} No. of Letters excluding vowels: Total = 5 {S, S, C, C, C,} After filling the even places with vowels:
Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word ⇒ mA = (No. of ways in which the 3 even places can be filled with the 2 vowels
Probability that the word formed using all the letters of the word «SUCCESS» such that the vowels will occupy odd places
Number of UnFavorable Choices = Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA ⇒ mAc = 5,040 − 2,160 OddsOdds in Favour of the words formed having all the vowels together = Number of Favourable Choices : Number of UnFavorable Choices = 2 : 3 Odds against the words formed having all the vowels together = Number of Unfavorable Choices : Number of Unfavourable Choices = 3 : 2 |
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| 05. |
In a random arrangement of the letters of the word �COMMERCE�. Find the probability that all the vowels come together Solution » Hide/Show Number of Letters/Characters in the word «COMMERCE» = 8 {C, O, M, M, E, R, C, E} ⇒ nL = 8 No. of Letters : of the first kind ⇒ No. of C’s = 2 ⇒ a = 2 of the second kind ⇒ No. of M’s = 2 ⇒ b = 2 In the experiment of testing for the number of words that can be formed using the letters of the word «COMMERCE» Total Number of Possible Choices = Number of words that can be formed using the 8 letters of the word «COMMERCE»
Let «A» be the event of all the vowels coming together For Event «A» No. of vowels = 3 {O, E, E} Considering all the vowels as a unit No. of letters = 6 {C, (O, E, E), M, M, R, C} No. of Letters : of the first kind ⇒ No. of C’s = 2 ⇒ a = 2 For the Group of Vowels, No. of Letters : of the first kind ⇒ No. of E’s = 2 ⇒ a = 2 Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word ⇒ mA = (No. of ways in which the 6 places can be filled with the letters (considering the vowels as a unit)) × (No. of ways in which the vowels can be interarranged
Probability that the word formed using all the letters of the word «COMMERCE» have all the vowels together
Number of UnFavorable Choices = Total Number of possible choices − Number of Favorable choices
⇒ mAc = n − mA ⇒ mAc = 2,880 OddsOdds in Favour of the words formed having all the vowels together = Number of Favourable Choices : Number of UnFavorable Choices = 2,160 × 2,880 Odds against the words formed having all the vowels together = Number of Unfavorable Choices : Number of Unfavourable Choices = 2,880 : 2,160 |
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| 06. |
The letters of the word ARRANGE are arranged at random. Find the probability that Solution » Hide/Show Number of Letters/Characters in the word «ARRANGE» = 7 {A, R, R, A, N, G, E} ⇒ nL = 7 No. of Letters : of the first kind ⇒ No. of A’s = 2 ⇒ a = 2 In the experiment of testing for the number of words that can be formed using the letters of the word «ARRANGE» Total Number of Possible Choices = Number of words that can be formed using the 7 letters of the word «ARRANGE»
Let «A», «B» «C» and «D» be the event of forming the words such that the two «R’s» are together, the two «R’s» are not together, the two «A’s» and the two «R’s» are together and the two «R’s» are together but the two «A’s» are not together respectively For Event «A» No. of vowels (a) = 5 {U, O, I, A, E} No. of other letters (Consonants) = 10 {N, C, P, Y, R, G, H, T, B, L} Considering the two «R’s» as a unit No. of Letters : of the first kind ⇒ No. of A’s = 2 ⇒ a1 = 2 Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word «ARRANGE» ⇒ mA = (No. of ways in which the 6 letters taking the two «R’s» as a unit can be
Probability that the word formed has the two «R’s» together
Number of UnFavorable Choices = Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA ⇒ mAc = 1,260 − 720 OddsOdds in Favour of the words formed having the two «R’s» together = Number of Favourable Choices : Number of UnFavorable Choices Odds against the words formed having all the vowels together = Number of Unfavorable Choices : Number of Unfavourable Choices For Event «B» Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word = Total number of words that can be formed ⇒ mB = n − mA ⇒ mB = mAc ⇒ mB = 540 Probability that the word formed does not have the two «R’s» together
Alternatively Probability that the word formed does not have the two «R’s» together ⇒ Probability of occurance of Event «B» = Probability of non-occurance of Event «A»
For Event «C» Considering the two «R’s» and the two «A’s» as a unit No. of Letters : which are all different = 4 {(A, A, R, R), N, G, E} ⇒ x = 4 In the letters considered as a unit, of the first kind ⇒ no. of A’s = 2 ⇒ a = 2 Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word «ARRANGE» ⇒ mC = (No. of ways in which the 4 letters taking the two «R’s» and the two «A’s»
Probability that the word formed has the two «R’s» adn two «A’s» together
For Event «D» With the two «R’s» as a unit and the two «A’s» as a unit No. of Letters : which are all different = 5 {(A, A), (R, R), N, G, E} ⇒ x = 5 With the two «R’s» as a unit No. of letters = 6 {A, (R, R), A, N, G, E} No. of Letters : of the first kind ⇒ No. of A’s = 2 ⇒ a = 2 Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word «ARRANGE» ⇒ mD = (No. of words that can be formed with the two «R’s» together ) − (No. of words
Probability that the word formed has the two «R’s» together and the two «A’s» not together
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| 07. |
If the letters of the word ASSASSIN� are written down at random in a row, the probability that no two s�s occur together is Solution » Hide/Show Number of Letters/Characters in the word «ASSASSIN» = 8 {A, S, S, A, S, S, I, N} ⇒ nL = 8 No. of Letters : of the first kind ⇒ No. of A’s = 2 ⇒ a = 2 of the second kind ⇒ No. of S’s = 4 ⇒ b = 4 Total Number of Possible Choices = No. of words that can be formed using the 11 letters of the word «ASSASSIN»
Let «G» be the event of forming the words such that no two «S’s» are together For Event «G» Placing the 4 S’s with gaps between them, {S_S_S_S} No. of Places for filling the other letters = 5 {_, S, _, S, _, S, _, S, _} Excluding the 4 «S’s», No. of Letters : in total = 4 {A, A, I, N} Number of Favourable/Favorable Choices = The number of words that can be formed using the letters of the word «ASSASSIN» such that no two «S’s» come together.
Probability that the word formed has no two «S’s» together
Number of UnFavorable Choices = Total Number of possible choices − Number of Favorable choices ⇒ mGc = n − mG ⇒ mAc = 1,680 − 60 OddsOdds in Favour of the words formed having the two «R’s» together = Number of Favourable Choices : Number of UnFavorable Choices Odds against the words formed having all the vowels together = Number of Unfavorable Choices : Number of Unfavourable Choices |
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| 08. |
Define the Event and identify the number of favourable choices in the following cases
Solution » Hide/Show
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| 09. |
Define the Event and identify the number of favourable choices in the following cases
Solution » Hide/Show
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| No. | Problems for Practice | ||||
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| 01. |
If the letters of the word �RANDOM� are arranged at random, then the probability that there are exactly 2 letters in between A and O is |
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| 02. |
The letters of the word TROPICAL are arranged in a row at random. Find the probability that no two vowels may come together. |
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| 03. |
The letters of the word WRESTLING are arranged in a row at random. The probability that the vowels may be in even places is |
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| 04. |
The probability that in a random arrangement of letters of the word COLLEGE the two E�s do not come together is |
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| 05. |
Define the Event and identify the number of favourable choices in the following cases
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Net Answers » Hide/Show
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Answers Here |
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