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
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
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 |
2 pages, 804 words
Emm as Dilemma In my investigation I am going to investigate the number of different arrangements of letters in a word. e. g. Tim Is one arrangement Mit Is another First I am going to investigate how many different arrangements in the name LUCY, which has no letters the same. LUCY LUC LYCUS LUC LC YU LC UY ULC Y UGLY ULC ULY C UCYLThere are 4 different letters and there are 24 different arrangements.
SAM SMA MSA MAS ASM AMS There are 3 different letters in this name and 6 different arrangements. JO OJ There are 2 different letters in this name and there are 2 different arrangements. UCL Y CUL Y UCL YUL C YL CU Y LUC CYL U C YUL CURL C ULY CLUB Table of Results Number of Letters Number of Different Arrangements 2 2 3 6 4 24 5 120 6 720 7 5040 From the table of results I have found out that a 2 letter word has 2 arrangements, and a 3 letter word has 6. Taking for example a 3 letter word, I have worked out that if we do 3 (the length of the word) x 2 = 6, the number of different arrangements. In a 4 letter word, to work out the amount of different arrangements you can do 4 x 3 x 2 = 24, or you can do 4! , which is called 4 factorial which is the same as 4 x 3 x 2. So, by using factorial (! ) I can predict that there will be 40320 different arrangements for an 8 letter word.
The formula for this is: n! = a Where n = the number of letters in the word and a = the number of different arrangements. Now I am going to investigate the number of different arrangements in a word with 2 letters repeated, 3 letters repeated and 4 letters repeated. EMMA A MME A MEM EXAM A EMM E AMM MME A MMA E MEM A MAME MEAN MEM 4 letter word, 2 letters repeated, 12 different arrangements. MUM MMU UMM 3 letter word, 2 letters repeated, 3 different arrangements. PQ MMM PMQMM PM MQM PMM MQ QP MMM QM PMM QM MPM QM MMP MPQMM MP MQM MPM MQ MQ PMM MQM PM MQM MP MMP QM MMQMP MMM PQ MMMQP MMP Q MMMQP 5 letter word, 3 letters repeated, 20 different arrangements. S MMM MSM M MMS M MMM S 4 letter word, 3 letters repeated, 4 different arrangements.
6 pages, 2702 words
The Term Paper on Cover Letter. What is it?
It is generally accepted practice to include a cover (or covering) letter, together with your resume and any other documentation that you forward to the employer as part of a job application. Your covering letter essentially provides an explanation of why you are communicating with the employer. Imagine a prospective employer’s confusion if they received your resume without a covering letter …
RRRR K RRRR RRRR RRRR K RRRR 5 letter word, 4 letters repeated, 5 different arrangements. Table of Results No of letters (n) No letters repeated Same letter repeated 2 x (p) Same letter repeated 3 x (p) Same letter repeated 4 x (p) Same letter repeated 5 x (p) 3 6 3 1 0 0 4 24 12 4 1 0 5 120 60 20 5 1 6 720 360 120 30 6 7 5040 2520 840 210 42 I have worked out that if you do say 5! = 120, to find out how many different arrangement in a 3 letter word it would be 5! Divided by 6 = 20, so, a 6 letter word with 4 letters repeated would be 6! Divided by 24 = 30, as you can see in the No letters repeated column these are the numbers we are dividing by: 2 letters the same = n! (21 = 2) 3 letters the same = n! (32 x 1 = 6) 4 letters the same = n! (43 x 21 = 24) 5 letters the same = n! (54 x 32 x 1 = 120) 6 letters the same = n! (65 x 43 x 21 = 720) From this I have worked out the formula to fine out the number of different arrangements: n! = the number of letters in the word p! = the number of letters the same Now I am going to investigate the number of different arrangement for words with 2 or more letters the same like, aabb, aaabb, or bbb aaa. This is a 4 letter word with 2 letters the same, there are 6 different arrangements: xxyyI am going to use the letters x and y (any letter) xxxyy xy xy yx xy xy yx yx yx yyxxThis is a 5 letter word xxxyy xxxyy xxyxy xxxx xyxyx xyxxy xyyxx yy xxx yxxxy yxyxx yxxyxThere are 10 different arrangements In the above example there are 3 x’s and 2 y’s As each letter has its own number of arrangements i. e. there were 6 beginning with x, and 4 beginning with y, I think that factorial has to be used again. As before, the original formula: n! = the number of letters in the word p! = the number of letters the same From this I have come up with a new formula The number of total letters factorial, divided by the number of x’s, y’s etc factor ised and multiplied.
3 pages, 1083 words
The Essay on Love: The Four Letter Word
Love, the four letter word that binds us all together. Whether it is the love of a significant other or your best friend, love is the glue to all relationships. Love is important as it give purpose and significance to bonds created with others. When you truly love another, the love for them is put above yourself. The main types of love are Companionship Love, Romantic or Sexual Love, Family Love, …
For the above example: A four letter word like aabb; this has 2 a’s and 2 b’s (2 x’s and 2 y’s) So: 12 x 34 (12) x (12) = 24 4 = 6 different arrangementsA five letter word like aaabb; this has 4 a’s and 1 b (4 x’s and 1 y) So: 12 x 34 x 5 (12 x 34) x (1) = 120 24 = 5 different arrangements A five letter words like abcde; this has 1 of each letter (no letters the same) So: 12 x 34 (11 x 11 x 11) = 24 1 = 24 different arrangementsA five letter word like aaabb; this has 3 a’s and 2 b’s (3 x’s and 2 y’s).
So: 12 x 34 x 5 (12 x 3) x (12) = 120 12 = 10 different arrangements This shows that my formula works: n! = the number of letters in the word x! y! = the number of repeated letters the same By Katherine Bond 11 W.
All Papers Are For Research And Reference Purposes Only. You must cite our web site as your source.
To expand upon what I already said in the comments, the probability of an event, $E$, occurring in an unbiased sample space, $S$, is given by the formula $$Pr(E) = frac{|E|}{|S|}$$
Note this only applies when every outcome in the sample space is equally likely to occur.
For this problem, our sample space is the set of ways in which the letters in the word ARRANGEMENTS can be arranged. From earlier example, we know that the number of arrangements of a word with $n_a$ copies of $A$, $n_b$ copies of $B$, …, $n_k$ copies of $K$ with $n=n_a+n_b+dots+n_k$ letters total will be given by the formula $$binom{n}{n_a,n_b,dots,n_k} = frac{n!}{n_a!n_b!cdots n_k!}$$
For our specific problem, we know then that ARRANGEMENTS has two $A$’s, two $E$’s, one $G$, one $M$, two $N$’s, two $R$’s, one $S$, and one $T$, and twelve letters total so $|S|=frac{12!}{2!2!1!1!2!2!1!1!}$ which simplifies as $=frac{12!}{2!2!2!2!}$ or even further as $=frac{12!}{(2!)^4}$. (Further simplifications are of course possible, but I will leave it like this so that it is clear how we found the numbers in the first place)
In part (a), it asks us what the probability is that having picked an arrangement of the word ARRANGEMENTS uniformly at random from all possible arrangements, that it has both $E$’s at the front of the word. To do this, we ask the question of how many arrangements of the word ARRANGEMENTS actually satisfies this property.
To count this, let us first guarantee the placement of the two $E$’s at the front. After this, we will place one of the possible arrangements of the remaining ten letters after the $E$’s. Answer the related question of how many ways we can arrange the remaining ten letters. Well, there are two $A$’s, one $G$, one $M$, two $N$’s, two $R$’s, one $S$, and one $T$ with ten letters total to arrange and place after the $E$’s for a total number of such arrangements as $frac{10!}{(2!)^3}$.
That was the count of how many there are. To get the probability, divide by the sample space size. $$Pr(E)=frac{|E|}{|S|}=dfrac{(frac{10!}{(2!)^3})}{(frac{12!}{(2!)^4})}=frac{10!2!}{12!}=frac{2}{12cdot 11}=frac{1}{66}$$
For part (b), we first count how many ways there are to arrange the letters in the word ARRANGEMENTS such that all consonants are together. To do this approach via multiplication principle.
To remind you, the multiplication principle states that if we want to count how many ways we can complete a task and we can describe every outcome uniquely via a sequence of steps with a specific number of options at each step which doesn’t rely on the choices that precede it, the total number of outcomes will be the product of the number of options for each step.
- Step one: Arrange the consonants by themselves. This can be accomplished in $binom{8}{1,1,2,2,1,1}=frac{8!}{(2!)^2}$ number of ways.
- Step two: Treating the group of consonants that have been arranged in step one as one big piece, lets call it $mathbb{X}$, arrange the vowels and $mathbb{X}$ together. We will arrange then $AAEEmathbb{X}$. This can be accomplished in $binom{5}{2,2,1}=frac{5!}{(2!)^2}$ number of ways.
Applying multiplication principle then, there are a total of:
$frac{8!}{(2!)^2}cdot frac{5!}{(2!)^2}=frac{8!5!}{(2!)^4}$ ways to do this
Applying the definition of probability in an unbiased sample space then, the probability is:
$dfrac{(frac{8!5!}{(2!)^4})}{(frac{12!}{(2!)^4})} = frac{8!5!}{12!}=frac{5cdot 4cdot 3cdot 2}{12cdot 11cdot 10cdot 9} = frac{1}{99}$