Find the number of permutations of the letters of the word permutations

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:

1. how many ways to arrange 2 letters word?
2. how many different ways to arrange 3 letters word?
3. how many distinguishable ways to arrange 4 letters word?
4. how many ways are there to order the letters LAKES?
5. how many distinguishable permutations of the letters of the word STATISTICS?
6. how many different ways can the letters of the word MATHEMATICS be arranged?
7. in how many ways can the letters of the word MATH be arranged?
8. find the number of distinct permutations of the word LAKES?
9. 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:

1. finding word permutation for words having distinct letters,
2. 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.

In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions.

Definition

Permutations are the different ways in which a collection of items can be arranged.

For example:

The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC.

Note that ABC and CBA are not same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.

The number of ways in which n things can be arranged, taken all at a time, ⁿP_n = n!, called ‘n factorial.’

Factorial Formula

Factorial of a number n is defined as the product of all the numbers from n to 1.

For example, the factorial of 5, 5! = 5*4*3*2*1 = 120.

Therefore, the number of ways in which the 3 letters can be arranged, taken all a time, is 3! = 3*2*1 = 6 ways.

Number of permutations of n things, taken r at a time, denoted by:

ⁿP_r = n! / (n-r)!

For example:

The different ways in which the 3 letters, taken 2 at a time, can be arranged is 3!/(3-2)! = 3!/1! = 6 ways.

Important Permutation Formulas

1! = 1

Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’.

Solution:

‘CHAIR’ contains 5 letters.

Therefore, the number of words that can be formed with these 5 letters = 5! = 5*4*3*2*1 = 120.

 
Problem 2: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘INDIA’.

Solution:

The word ‘INDIA’ contains 5 letters and ‘I’ comes twice.

When a letter occurs more than once in a word, we divide the factorial of the number of all letters in the word by the number of occurrences of each letter.

Therefore, the number of words formed by ‘INDIA’ = 5!/2! = 60.

 
Problem 3: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘SWIMMING?

Solution:

The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice.

Therefore, the number of words formed by this word = 8! / (2!*2!) = 10080.

 
Problem 4: How many different words can be formed with the letters of the word ‘SUPER’ such that the vowels always come together?

Solution:

The word ‘SUPER’ contains 5 letters.

In order to find the number of permutations that can be formed where the two vowels U and E come together.

In these cases, we group the letters that should come together and consider that group as one letter.

So, the letters are S,P,R, (UE). Now the number of words are 4.

Therefore, the number of ways in which 4 letters can be arranged is 4!

In U and E, the number of ways in which U and E can be arranged is 2!

Hence, the total number of ways in which the letters of the ‘SUPER’ can be arranged such that vowels are always together are 4! * 2! = 48 ways.

 
Problem 5: Find the number of different words that can be formed with the letters of the word ‘BUTTER’ so that the vowels are always together.

Solution:

The word ‘BUTTER’ contains 6 letters.

The letters U and E should always come together. So the letters are B, T, T, R, (UE).

Number of ways in which the letters above can be arranged = 5!/2! = 60 (since the letter ‘T’ is repeated twice).

Number of ways in which U and E can be arranged = 2! = 2 ways

Solution:

The word ‘REMAINS’ has 7 letters.

There are 4 consonants and 3 vowels in it.

Writing in the following way makes it easier to solve these type of questions.

(1) (2) (3) (4) (5) (6) (7)

No. of ways 3 vowels can occur in 4 different places = ⁴P₃ = 24 ways.

After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = ⁴P₄ = 4! = 24 ways.

Therefore, total number of permutations possible = 24*24 = 576 ways.

Combinations

Definition

The different selections possible from a collection of items are called combinations.

For example:

The different selections possible from the alphabets A, B, C, taken 2 at a time, are AB, BC and CA.

It does not matter whether we select A after B or B after A. The order of selection is not important in combinations.

To find the number of combinations possible from a given group of items n, taken r at a time, the formula, denoted by ⁿC_r is

ⁿC_r = n! / [r! * (n-r)!]

For example, verifying the above example, the different selections possible from the alphabets A, B, C, taken two at a time are

³C₂ = 3! / (2! * (3-2)!) = 3 possible selections (i.e., AB, BC, CA)

Important Combination formulas

ⁿC_n = 1

ⁿC₀ = 1

ⁿC₁ = n

ⁿC_r = ⁿC_(n-r)

The number of selections possible with A, B, C, taken all at a time is ³C₃ = 1 (i.e. ABC)

Solved examples of Combination

Let us take a look at some examples to understand how Combinations work:

 
Problem 1: In how many ways can a committee of 1 man and 3 women can be formed from a group of 3 men and 4 women?

Solution:

No. of ways 1 man can be selected from a group of 3 men = ³C₁ = 3! / 1!*(3-1)! = 3 ways.

No. of ways 3 women can be selected from a group of 4 women = ⁴C₃ = 4! / (3!*1!) = 4 ways.

Problem 2: Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls.

Solution:

Selecting at least 3 black balls from a set of 5 black balls in a total selection of 5 balls can be

3 B and 2 R

4 B and 1 R and

5 B and 0 R balls.

Therefore, our solution expression looks like this.
⁵C₃ * ³C₂ + ⁵C₄ * ³C₁ + ⁵C₅ * ³C₀ = 46 ways .

 
Problem 3: How many 4 digit numbers that are divisible by 10 can be formed from the numbers 3, 5, 7, 8, 9, 0 such that no number repeats?

Solution:

If a number is divisible by 10, its units place should contain a 0.
_ _ _ 0

After 0 is placed in the units place, the tens place can be filled with any of the other 5 digits.

Selecting one digit out of 5 digits can be done in ⁵C₁ = 5 ways.

After filling the tens place, we are left with 4 digits. Selecting 1 digit out of 4 digits can be done in ⁴C₁ = 4 ways.

After filling the hundreds place, the thousands place can be filled in ³C₁ = 3 ways.

Therefore, the total combinations possible = 5*4*3 = 60.

Permutations and Combinations Quiz

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Problem 1: Click here

Answer 1: Click here

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Problem 2: Click here

Answer 2: Click here

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Problem 3: Click here

Answer 3: Click here

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In mathematics, permutation relates to the method of organizing all the members of a group into some series or design. In further terms, if the group is already completed, then the redirecting of its components is called the method of permuting. Permutations take place, in better or slightly effective methods, in almost every district of mathematics. They usually happen when different direction on detailed restricted sites is monitored.

Permutation formula

It is the separate arrangements of a supplied numeral of associates taken one by one, or some, or all at a time. For instance, if we have two elements A and B, then there are two possible interpretations, AB and BA.

An integer of permutations when ‘r’ elements are arranged out of a total of ‘n’ elements is,

ⁿP_r = n! / (n – r)!. 

For example, 

Let n = 2 (A and B) and r = 1 (All permutations of size 1). The answer is 2!/(2 – 1)! = 2. The two permutations are AB, and BA.

Explanation of Permutation formula

A permutation is a kind of arrangement that shows how to permute. If there are three separate integers 1, 2, and 3, and if somebody is interested to permute the integers taking 2 at a point, it offers (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That is it can be performed in 6 ways.

Here, (1, 2) and (2, 1) are separate. Again, if these 3 integers shall be set enduring all at a time, then the arrangements will be (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) i.e. in 6 ways.

In known, n separate items can be selected accepting r (r < n) at a time in n(n – 1)(n – 2) … (n – r + 1) ways. In particular, the first item can be any of the n items. Now, after selecting the first item, the second item will be any of the remaining n – 1 thing. Similarly, the third item can be any of the remaining n – 2 things. Alike, the r^th item can be any of the remaining n – (r – 1) things.

Therefore, the total numeral of permutations of n separate items taking r at a time is n(n – 1)(n – 2) … [n – (r – 1)] which is noted as ⁿP_r. Or, in other words,

ⁿP_r = n!/(n – r)!

Sample Problems

Question 1: What are the types of permutations?

Solution:

The permutation of a collection of things or components in order relies on three conditions:

1. When recurrence of essences is not allowed
2. When recurrence of essences is allowed
3. When the components of a group are not different

Question 2: Calculate the number of permutations of n = 5 and r = 2.

Solution:

Given,

n = 5

r = 2

Using the formula given above:

Permutation: ⁿP_r = (n!) / (n – r)!

= (5!) / (5 – 2)!

= 5! / 3! = (5 × 4 × 3! )/ 3!

= 20

Question 3: How many 3 letter phrases with or without purpose can be created out of the letters of the word POEM when repetition of letters is not permitted?

Solution:

Here n = 4, as the word POEM has 4 letters. Since we have to create 3 letter words with or without meaning and without repetition, therefore total permutations possible are:

⇒ P(n, r) = 4!/(4 − 3)!

= 4 × 3 × 2 × 1/1

= 24

Question 4: How many 4 letter phrases with or without purpose can be created out of the letters of the word KANHA when repetition of words is permitted?

Solution:

The number of letters, in this case, is 5, as the word KANHA has 5 alphabets.

And r = 4, as a 4-letter term has to be selected.

Thus, the permutation will be:

Permutation (when repetition is permitted) = 5⁴= 625

Question 5: It is required to place 4 men and 3 women in a row so that the women entertain the even positions. How many such configurations are feasible?

Solution: 

We are given that there are 4 men and 3 women.

i.e. there are 7 positions.

The even positions are: 2nd, 4th, and the 6th places

These three places can be occupied by 3 women in P(3, 3) ways = 3!  

= 3 × 2 × 1  

= 6 ways

The remaining 4 positions can be occupied by 4 men in P(4, 4) = 4!  

= 4 × 3 × 2 × 1  

= 24 ways

Therefore, by the Fundamental Counting Principle,  

Total number of ways of seating arrangements = 24 ×  6 

= 144

Question 6: Find the number of phrases, with or without meaning, that can be comprised with the letters of the word ‘TABLE’.

Solution:

‘TABLE’ contains 5 letters.

Thus, the numeral of phrases that can be formed with these 5 letters = 5! = 5 × 4 × 3 × 2 × 1 = 120.

Question 7: Find the number of permutations of the letters of the phrase subject such that the vowels consistently appear in odd positions.

Solution:

The word ‘SUBJECT’ has 7 letters.

There are 6 consonants and 1 vowels in it.

No. of ways 1 vowels can occur in 7 different places = ⁷P₁ = 7 ways.

After 1 vowels take 1 place, no. of ways 6 consonants can take 6 places = ⁶P₆ = 6! = 720 ways.

Therefore, total number of permutations possible = 720 × 720 = 518,400 ways.

The word PERMUTATION has eleven letters, of which six are consonants and five are vowels. We will first arrange the consonants, then place the vowels. We will consider two cases, depending on whether the Ts are separated or together in the arrangement of the consonants.

Case 1: We arrange the six consonants so that the Ts are separated.

There are $4!$ of the consonants P, R, M, N. This creates five spaces in which to place the Ts.
$$square C_1 square C_2 square C_3 square C_4 square$$
To ensure the Ts are separated, we must choose two of these five spaces in which to place the $T$s, which can be done in $binom{5}{2}$ ways. We now have seven spaces in which to place the vowels.
$$square C_1 square C_2 square C_3 square C_4 square C_5 square C_6 square$$
To ensure the vowels are separated, we must choose five of these seven spaces in which to place a vowel, which can be done in $binom{7}{5}$ ways. The five vowels can be arranged in the selected spaces in $5!$ ways. Hence, there are
$$4!binom{5}{2}binom{7}{5}5!$$
such arrangements.

Case 2: We arrange the six consonants so that the Ts are together.

We have five distinct objects to arrange: P, R, M, N, TT. They can be arranged in $5!$ ways. Now that we have arranged the six consonants, we again have seven spaces in which to place the five vowels. To ensure the Ts are separated, we must place one of the vowels between the two Ts. To ensure the vowels are separated, we must also place a vowel in four of the remaining six spaces, which can be done in $binom{6}{4}$ ways. We can arrange the five vowels in the five selected spaces in $5!$ ways. Hence, there are
$$5!binom{6}{4}5!$$
such arrangements.

Total: Since these cases are mutually exclusive and exhaustive, the number of admissible arrangements is
$$4!binom{5}{2}binom{7}{5}5! + 5!binom{6}{4}5!$$

In permutation with repetition, the elements are allowed to repeat. Permutation with Repetition is the simplest to determine. In this article, we will learn about, Permutation with Repetition explained with proof, formulae and solved examples, the definition of permutation, types, circular and uses of permutations.

What is Permutation?

The distinct methods of arranging a set of things into a sequential order are termed permutation.

For example:

• Ordering characters, digits, symbols, alphabets, letters, and colors.
• Selecting first, second and third positions for the winners.

Mathematically, the permutation is associated with the act of arranging all the data of a set into some sequence or order. Permutations come into account in more or less almost every domain of mathematics. For this, we use the standard permutation formula. There are distinct properties of permutations apart from the standard formula.

Learn the various concepts of the Binomial Theorem here.

The Permutation with Repetition is the simplest to determine. Consider when a piece has n different types and one has r choices each time then the permutations is defined by: n × n × … (r times). This implies there are n possibilities for the first selection, followed by n possibilities for the second selection, and so on, multiplying each time.

Permutation with Repetition Formula

The number of permutations of n objects, where p objects are of one kind, q objects are of another kind and the rest, if any, are of a different kind is (begin{align*}frac{_nP_r}{p! q!}end{align*})

Check out this article on Arithmetic Mean.

How do you Solve Permutation with Repetition?

Consider the word TEETH. There are 2 E’s in the word. Both E’s are identical, and it does not matter in which order we write these 2 E’s since they are the same. In other words, if we exchange ‘E’ for ‘E’, we still spell TEETH. The same is true for the T’s since there are 2 T’s in the word TEETH as well. In how many ways can we arrange the letters in the word TEETH?

We must account for the fact that these 2 E’s are identical and that the 2 T’s are identical. We do this using the formula:

(frac{_nP_r}{x_1! x_2!}), where x is the number of times a letter is repeated.

(begin{matrix}frac{_nP_r}{x_1! x_2!} &= frac{_5P_5}{2!2!}\frac{_5P_5}{2!2!} &= frac{5 times 4 times 3 times 2 times 1}{2 times 1 times 2 times 1}\frac{_5P_5}{2!2!} &= frac{120}{4}\frac{_5P_5}{2!2!} &= 30end{matrix})

Also, read about Arithmetic progressions with this article.᠎

Circulation Permutations with Repetition

The formula for Circulation Permutations with Repetition for n elements is = (frac{n!}{n} = (n-1))

Let us determine the number of distinguishable permutations of the letters ELEMENT.

Suppose we make all the letters different by labelling the letters as follows.

(E_1LE_2ME_3NT)

Now, all the letters are different from each other. In this case, there are (7-1)! = 6! different circular permutations possible.

Let us consider one such case of circular permutation, where the elements are arranged in the order below:

(LE_1ME_2NE_3T)

We can form new arrangements of permutations from this arrangement by only moving the E’s. Clearly, there are 3! or 6 such arrangements. We list them below.

(begin{matrix}LE_1ME_2NE_3\LE_1ME_3NE_2\LE_2ME_1NE_3T\LE_2ME_3NE_1T\LE_3ME_2NE_1T\LE_3ME_1NE_2Tend{matrix})

Because the E’s are not different, there is only one arrangement LEMENET and not six. This is true for every permutation.

Also, learn about Mean Deviation.

Solved Examples on Permutation with Repetition

1. Find the number of permutations of the letters of the word MICROSOFT.

A.  The word MICROSOFT consists of 9 letters, in which the letter ‘O’ is repeated two times. Therefore, the number of permutations of the letters of the word MICROSOFT = ({9!over{2!}}) = 181440.

2. How many arrangements can be made, with the letters of the word CALCULATOR? In how many of these arrangements, vowels occur together?

A.  The word CALCULATOR consists of 10 letters, in which ‘C is repeated two times, ‘A’ is repeated two times, ‘L’ is repeated two times and the rest all are different.
Therefore, the number of permutations of the letters of the word CALCULATOR = ({10!over{2!2!2!}}) = 453600
The word CALCULATOR consists of 4 vowels A, U, A, 0. Let us consider them as a single letter say P
Therefore, now we have 7 letters P, C, L, C, L, T, R in which “C” is repeated two times, L is repeated two times. The number of arrangement of these 7 letters is 7!
is given ({7!over{2!2!}}) = 1260.

After this is done, 4 vowels (in which ‘A’ is repeated 2 times) can be arranged in ({4!over{2!}}) = 12 ways
Therefore, the number of arrangements of the letters of the word CALCULATOR in which vowels are together = 1260 x 12 = 15120.

3. John owns six-coloured pairs of shoes (two red, two blue and two black). He wants to put all these pairs of shoes on the shoe rack. How many different arrangements of shoes are possible?

A.  The total number of pair of shoes = n = 6
Number of red shoes = p = 2
Number of blue shoes = q = 2
Number of back shoes = r = 2
This is an example of permutation with repetition because the elements are repeated and their order is important.
Put the above values in the formula below to get the number of permutations:
(begin{matrix}PR^{p,q,r} _n = frac {n!}{p!q!r!}\PR^{2,2,2} _n = frac {6!}{2!2!2!}\= frac{720}{8}\= 90\end{matrix})
Hence, shoes can be arranged on the shoe rack in 90 ways.

4. A person has to choose three digits from the set of the following seven numbers to make a three-digit number.{1, 2, 3, 4, 5, 6, 7} How many different arrangements of the digits are possible?

A.  A three-digit number can have 2 or three identical numbers. Similarly, in a number, the order of digits is important.It is given that the person can select 3 digits from the set of 7 numbers. Hence, n = 7 and k = 3. Substitute these values in the formula below to get the number of possible arrangements.
P(n,k) = (n^k)
P(7,3) = (7^3)
=343
Hence, 343 different arrangements are possible.

5. If there are 4 chocolate chips, 2 oatmeal, and 2 double chocolate cookies in a box, in how many different orders is it possible to eat all of these cookies?

A:Total number of cookies = 8, so n = 8.
We want to eat all of the cookies; therefore, we are choosing 8 objects at a time. Therefore, r = 8.
In the group of 8 cookies, there are 4 chocolate chips, 2 oatmeal, and 2 double chocolate cookies. We are eating a group of cookies with flavours that repeat.
(begin{matrix}
x_1 = 4,\
x_2 = 2,\
x_3 = 2
end{matrix})

(begin{matrix}
{^nP_rover{x_1!x_2!x_3!}} = {^8P_8over{4!2!2!}}\
{^8P_8over{4!2!2!}} = {8times7times6times5times4times3times2times1over{4times3times2times1times2times1times2times1}} = {40320over{96}}\
{^8P_8over{4!2!2!}} = 420
end{matrix})

6. In how many ways can the alphabets of the word EXCELLENT be arranged?

A. Total number of elements in the word = n = 9
E is repeated three times, hence p = 3
L is repeated 2 times, hence q = 2
Substitute these values in the formula below to get the number of ways in which the letters of this word can be arranged:
(begin{matrix}PR^{p,q} _n = frac {n!}{p!q!}\PR^{3,2} _n = frac {9!}{3!2!}\PR^{3,2} _n = frac {9!}{3!2!}\PR^(3,2)_n = frac{362880}{12}\= 30240end{matrix})
Hence, the letters in the word EXCELLENT can be arranged in 30240 ways.

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Permutation with Repetition FAQs