# Permutations and Combinations Problems | GMAT GRE Maths Tutorial

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

## Permutations

#### 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, nPn = 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:

nPr = 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

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 = 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 nCr is

nCr = 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

nCn = 1

nC0 = 1

nC1 = n

nCr = 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

Try these practice problems.

Start here | Success stories | Reality check | Knowledgebase | Scholarships | Services ### 31 thoughts on “Permutations and Combinations Problems | GMAT GRE Maths Tutorial”

1. 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

2. 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!

• as per my understanding there are two “T ‘s”, in order that it should be placed in the begining there are 2!ways
considering (TT) as a single element we are left with 9 letters (i.e) M,A,H,E,M,A,I,C,S and the number of ways it can placed is 9!/(2!*2!) …..so the final answer is 9!/(2!)

3. 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

• I don’t understand. Does the order in which you draw the white balls matter? If so, I figure 6 ways. If all that matters is whether or not your three balls are comprised of 2 whites or 1 white then the answer is 2. Please explain. Maybe the question is not fully developed..? Or maybe my answers aren’t fully developed! HELP!

• There are two ways to choose three balls and have at least one be white. You can either choose one white and two red or two white and one red. Arrangement doesn’t matter. The answer is two.

Try checking your answer by attempting to write out 9 or 12 ways of choosing the balls.

• CORRECT.
Only 3 balls can be taken so,
Posibility A – 1 x w and 2 red taken – 1st white taken, then the possibility of 2 red from 3 balls is – 3C2 = 3
Posibility B – 1 x w and 2 red – 2nd white taken, then the possibility of 2 red from 3 balls is – 3C2 = 3
Posibility C – 2 x w and 1 red – both white taken, then the possibility of 1 red from 3 balls is – 3C3 = 3

So the total possibilities is 3+3+3 = 9
Hope all got this. Can help if anyone can’t

• 2c1x3c2+2c2x3c1 =6+ 3 = 9

4. 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

• One ball must be white, so we keep it fixed. That leaves us with 4 choices for the remaining 2 slots.
So, 4 ways to choose the first ball and 3 ways to choose the second one, 3*4= 12
Isn’t that an easier approach?

5. 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.

• you have 3 shirts and 2 pants you can wear one pants and one shirt at a time so it will be 3^C1*2^C1 = 3×2 = 6.

6. 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?