Are you desperately looking for practice questions on the GMAT club forum? How many questions did you find for the topic you were looking for? Are they any good?

Well, if you are scavenging for good practice questions you’ve just hit a gold mine! Problem solving comprises roughly 50% of the total questions in the GMAT quant section and in this blog, we will show you how to solve GMAT permutation and combination questions through some quality practice questions and their detailed explanations. Let’s go!

#### IN THIS BLOG:

1. GMAT Permutation and Combination Problem 1

2. GMAT Permutation and Combination Problem 2

3. GMAT Permutation and Combination Problem 3

4. GMAT Permutation and Combination Problem 4

5. GMAT Permutation and Combination Problem 5

6. GMAT Permutation and Combination Problem 6

7. GMAT Permutation and Combination Problem 7

8. GMAT Permutation and Combination Problem 8

9. GMAT Permutation and Combination Problem 9

10. GMAT Permutation and Combination Problem 10

11. GMAT Permutation and Combination Problem 11

12. GMAT Permutation and Combination Problem 12

13. GMAT Permutation and Combination Problem 13

## GMAT Permutation and Combination Problem 1:

How many 3-digit numbers can be formed out of the digits 1, 2, 3, 4, and 5?

**Solution:** Forming numbers requires an ordered selection. Hence, the answer will be ^{5}P_{3}

## GMAT Permutation and Combination Problem 2:

In how many ways can the 7 letters M, N, 0, P, Q, R, S be arranged so that P and Q occupy continuous positions?

**Solution:** For arranging the 7 letters keeping P and Q always together we have to view P and Q as one letter. Let this be denoted by PQ.

Then, we have to arrange the letters M, N, O, PQ, R, and S in a linear arrangement. Here, it is like arranging 6 letters in 6 places (since 2 letters are counted as one). This can be done in 6! ways.

However, the solution is not complete at this point in time since in the count of 6! the internal arrangement between P and O is neglected. This can be done in 2! ways. Hence, the required answer is 6! X 2!.

Task for you: What would happen if the letters P,Q, and R are to be together?

(Answer: 5! x 3!)

And what if P and Q are never together?

(Answer will be given by the formula: Total number of ways – Number of ways they are always together)

## GMAT Permutation and Combination Problem 3:

Of the different words that can be formed from the letters of the words BEGINS how many begin with B and end with S?

**Solution:** B & S are fixed at the start and the end positions. Hence, we have to arrange E, G, I, and N amongst themselves. This can be done in 4! ways.

Task for you: What will be the number of words that can be formed with the letters of the word BEGINS which have B and S at the extreme positions?

(Answer: 4! x 2!)

## GMAT Permutation and Combination Problem 4:

In how many ways can the letters of the word VALEDICTORY be arranged, so that all the vowels are adjacent to each other?

**Solution:** There are 4 vowels and 7 consonants in Valedictory. If these vowels have to be kept together, we have to consider AEIO as one letter. Then the problem transforms itself into arranging 8 letters amongst themselves (8! ways). Besides, we have to look at the internal arrangement of the 4 vowels amongst themselves. (4! ways)

Hence Answer = 8! x 4!.

## GMAT Permutation and Combination Problem 5:

If there are two kinds of hats, red and blue, and at least 5 of each kind, in how many ways can the hats be put in each of 5 different boxes?

**Solution:** The significance of at least 5 hats of each kind is that while putting a hat in each box, we have the option of putting either a red or a blue hat. (If this was not given, there would have been uncertainty in the number of possibilities of putting a hat in a box.)

Thus in this question for every task of putting a hat in a box, we have the possibility of either putting a red hat or a blue hat. The solution can then be looked at as: there are 5 tasks each of which can be done in 2 ways. Through the MNP rule, we have the total number of ways = 2^{5} (Answer).

## GMAT Permutation and Combination Problem 6:

In how many ways can 4 Indians and 4 Nepalese people be seated around a round table so that no -two Indians are in adjacent positions?

**Solution:** If we first put 4 Indians around the round table and we can do this in 3! ways.

Once the 4 Indians are placed around the round table, we have to place the four Nepalese around the same round table. Now, since the Indians are already placed we can do this in 4! ways (as the starting point is defined when we put the Indians. Try to visualize this around a circle for placing 2 Indians and 2 Nepalese.

Hence, Answer = 3! x 4!

## GMAT Permutation and Combination Problem 7:

If there are 11 players to be selected from a team of 16, in how many ways can this be done?

Solution. ^{16}C_{11}

Have you taken the GMAT before?

## GMAT Permutation and Combination Problem 8:

In how many ways can 18 identical white and 16 identical black balls be arranged in a row so that no two black balls are together?

**Solution:** When 18 identical white balls are put in a straight line, there will be 19 spaces created. Thus 16 black balls will have 19 places to fill in. This will give an answer of ^{19}C_{16}. (Since the balls are identical the arrangement is not important.)

## GMAT Permutation and Combination Problem 9:

A mother with 7 children takes three at a time to a cinema. She goes with every group of three that she can form. How many times can she go to the cinema with distinct groups of three children?

**Solution:** She will be able to do this as many times as she can form a set of three distinct children from amongst the seven children. This essentially means that the answer is the number of selections of 3 people out of 7 that can be done. Hence, Answer = ^{7}C_{3}

## GMAT Permutation and Combination Problem 10:

Based on the above question, how many times will an individual child go to the cinema with her before a group is repeated?

**Solution:** This can be viewed as: The child for whom we are trying to calculate the number of ways is already selected. Then, we have to select 2 more children from amongst the remaining 6 to complete the group. This can be done in ^{6}C_{2} ways.

## GMAT Permutation and Combination Problem 11:

How many different sums can be formed with the following coins?

5 rupee, 1 rupee, 50 paisa, 25 paisa, 10 paisa and 1 paisa

**Solution:** A distinct sum will be formed by selecting either 1 or 2 or 3 or 4 or 5 or all 6 coins.

But from the formula, we have the answer to this as 2^{6}– 1.

[Task for you: How many different sums can be formed with the following coins?

5 rupee, 1 rupee, 50 paisa, 25 paisa, 10 paisa, 3 paisa, 2 paisa and 1 paisa?

[Hint: You will have to subtract some values for double-counted sums.]

## GMAT Permutation and Combination Problem 12:

A train is going from Mumbai to Pune and makes 5 stops on the way. 3 persons enter the train during the journey with 3 different tickets. How many different sets of tickets may they have had?

**Solution:** Since the 3 persons are entering during the journey they could have entered at the:

1^{st} station (from where they could have bought tickets for the 2^{nd}, 3^{rd}, 4^{th} or 5^{th} stations or for Pune -a total of 5 tickets.)

2^{nd} station (from where they could have bought tickets for the 3^{rd}, 4^{th} or 5^{th} stations or for Pune – a total of 4 tickets.)

3^{rd} station (from where they could have bought tickets for the 4^{th} or 5^{th} stations or for Pune – total of 3 tickets.)

4^{th} station (from where they could have bought tickets for the 5^{th} station or for Pune – total of 2 tickets.)

5^{th} station (from where they could have bought a ticket for Pune – total of 1 ticket.)

Thus, we can see that there are a total of 5 + 4 + 3 + 2 + 1 = 15 tickets available out of which 3 tickets were selected. This can be done in ^{15}C_{3}, ways (Answer).

## GMAT Permutation and Combination Problem 13:

Find the number of diagonals and triangles formed in a decagon.

**Solution:** A decagon has 10 vertices. A line is formed by selecting any two of the ten vertices. This can be done in ^{10}C_{2} ways. However, these ^{10}C_{2} lines also count the sides of the decagon.

Thus, the number of diagonals in a decagon is given by:

^{10}C_{2 }– 10

Triangles are formed by selecting any three of the ten vertices of the decagon. This can be done in ^{10}C_{3 }ways.

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