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

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, ^{n}P_{n} = n!, called ‘n factorial.’

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:

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.

1! = 1

0! = 1

Let us take a look at some examples:

**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

Therefore, total number of permutations possible = 60*2 = 120 ways.

**Problem 6:** Find the number of permutations of the letters of the word ‘REMAINS’ such that the vowels always occur in odd places.

__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 = ^{4}P_{3 }= 24 ways.

After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = ^{4}P_{4} = 4! = 24 ways.

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

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 ^{n}C_{r} is

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

^{3}C_{2 }= 3! / (2! * (3-2)!) = 3 possible selections (i.e., AB, BC, CA)

^{n}C_{n} = 1

^{n}C_{0 }= 1

^{n}C_{1 }= n

^{n}C_{r }= ^{n}C_{(n-r)}

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

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 = ^{3}C_{1 }= 3! / 1!*(3-1)! = 3 ways.

No. of ways 3 women can be selected from a group of 4 women = ^{4}C_{3 }= 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.

^{5}C_{3 }* ^{3}C_{2 } + ^{5}C_{4} * ^{3}C_{1} + ^{5}C_{5 }* ^{3}C_{0} = 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 ^{5}C_{1 }= 5 ways.

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

After filling the hundreds place, the thousands place can be filled in ^{3}C_{1 }= 3 ways.

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

Try these practice problems.

Problem 1

Solve the following.

i) ^{30}P_{2 }

ii) ^{30}C_{2}

A. 870, 435

B. 435, 870

C. 870, 470

D. 435, 835

Answer 1

A

**Explanation:**

^{30}P_{2} = 30! / 28! = 30*29*28! / 28! = 30*29 = 870.

^{30}C_{2 }= 30! / (2!*28!) = 435.

Problem 2

How many different possible permutations can be made from the word ‘BULLET’ such that the vowels are never together?

A. 360

B. 120

C. 480

D. 240

Answer 2

D.

**Explanation**:

The word ‘BULLET’ contains 6 letters of which 1 letter occurs twice = 6! / 2! = 360

No. of permutations possible with vowels always together = 5! * 2! / 2! = 120

No. of permutations possible with vowels never together = 360-120 = 240.

Problem 3

In how many ways can a selection of 3 men and 2 women can be made from a group of 5 men and 5 women ?

A. 10

B. 20

C. 30

D. 100

Answer 3

D.

**Explanation**:

^{5}C_{3} * ^{5}C_{2} = 100

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Got queries unrelated to this article? Post them on our General Queries Page.## 21 Comments

How many six letter word can be formed with the letters of the word ‘policy’ such that the vowel can only occur even place?

6 ways for arranging 2 vowels * 24 ways for arranging 4 consonants =144

how many different arrangements can be made by using all the letters of the word MATHEMATICS ? How many of them begin with C? How many of them begin with T? In how many of them consonants will occur together?

1. 11!/2!*2!*2!

2.10!/2!*2!*2!

3.(2!*10!)/2!*2!*2!

4.(4!*7!)/2!*2!*2!

Correct me if its worng!

How the answer of 3 is (2!*10!)/(2!*2!*2!). As per my understanding it should be (10!)/(2!*2!). The answer is same but the approach should be different.

Ans of 4 should be (5!*7!)/2!*21*2!

A bag contains 2 white and 3 red balls. In how many ways can 3 balls be chosen if,at least one ball must be white?

2C2 * 3C1 + 2C1 * 3C2

2c1*3c2+2c2*3c1=12

The ans is 9 not 12

How many words can be formed of the letter TOPOLOGY in which two vowels are never together.

(8!/3!)-(7!*3p2)

Correct me if it’s wrong!

you have approached this same as PROBLEM 2. But 3 O’s make your situation complicated. Try finding the answer as in example 6 from permutaion. Answer should be 2400. (5! * 6C3 ). (NOTE: here you have to use 6C3 instead of 6P3)

i think it should be 8!/3! – 6!*3!/3!

= 6720-720= 6000

The correct answer for this 2400.

First take all the consonants together : TPLGY. These can be arranged in following ways : 5! = 120

Now make an arrangement like this : _T_P_L_G_Y_. This ensure no two vowels will be together. The number of ways you can do this

P(6,3)/3! = 20 (divide by 3! because vowels are ‘0’,’0′ and ‘0’, three times repetition) .

Now just multiply these two possibilities: 120 * 20 = 2400

Hiya, Have a doubt. Combinations problem 3 – “Therefore, the total combinations possible = 5*4*3 = 60.”

Why are we multiplying the combinations and not adding them up to get the total numbers?

these 3 events occur simultaneously, and independently. try finding the combination of dress code u can come up with 3 shirt and 2 pants. in these situation multiplication gives you total possible combination.

How many different 4 letters words can be formed from the word examination

A coustomer forgets a four digit ATM code.He remembers that this code consist of digits 3,5,6,9. Find the maximum no. of trails he needed to make to obtain the code?

With the word TOMORROW, how many different words can you form such that the vowels in the words should not come together?

There are 10 bus service from place A to B .In how many ways can a man go from A to B and return by different bus?