# Modern Math - Permutation & Combination - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Modern Math - Permutation & Combination. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2019 QADI | Modern Math - Permutation & Combination**

A bag contains marbles of three colours-red, blue and green. There are 8 blue marbles in the bag. There are two additional statement of facts available:

- If we pull out marbles from the bag at random, to guarantee that we have at least 3 green marbles, we need to extract 17 marbles.
- If we pull out marbles from the bag at random, to guarantee that we have at least 2 red marbles, we need to extract 19 marbles.

Which of the two statements above, alone or in combination shall be sufficient to answer the question "how many green marbles are there in the bag"?

- A.
Both statements taken together are suﬃcient to answer the question, but neither statement alone is suﬃcient.

- B.
Each statement alone is suﬃcient to answer the question.

- C.
Statements 1 and 2 together are not suﬃcient, and additional data is needed to answer the question.

- D.
Statement 2 alone is suﬃcient, but statement 1 alone is not suﬃcient to answer the question.

- E.
Statement 1 alone is suﬃcient, but statement 2 alone is not suﬃcient to answer the question.

Answer: Option D

**Explanation** :

We know that in all there are 8 blue marbles.

Let us first look at statement I.

As per statement I if we are to pull out 17 marbles from the bag, we will ensure that there are at least 3 greeen marbles. Now out of 17 marbles removed, 8 are blue .So from the 9 marbles that are removed, if at least 3 are green, it would mean maximum possible no of red marbles removed are 9-3 or 6.Which means that the red marbles in the bag are 6.However, this statement alone gives us no information about the no of green marbles. Hence statement I alone is not sufficient to answer the question.

Let us next look at statement II.

As per this statement, if we are to pull out 19 marbles from the bag, we would have at least 2 red marbles. Out of the 19 marbles removed, suppose 8 are blue. Now out of the remaining 11 marbles removed, if we have at least 2 red marbles, it would mean that the maximum possible no of green marbles removed is 9.This means that in all there are 9 green marbles in the bag .

Hence statement 2 alone is sufficient to answer the question.

Hence, option (d).

Workspace:

**IIFT 2019 QA | Modern Math - Permutation & Combination**

Four couples are to be seated in a circular table such that each couple sits together. In how many ways they can sit such that two males sit to the right of their female partners and the other two males sit to the left of their female partners?

- A.
144

- B.
288

- C.
1440

- D.
720

Answer: Option B

**Explanation** :

First of all, we will have to select 2 couples where the male sits to the right of the female in ^{4}C_{2} = 6 ways (and the other two couples select them automatically)

Now, we can fix anyone of the couple in 4 ways and in these couples, females may be seated to the left/right of the males

Total ways of fixing a couple = 8

The other three couples can be arranged in 3! ways = 6 ways

∴ Total number of ways = 6 × 8 × 8 = 288

Hence, option (b).

Workspace:

**IIFT 2018 QA | Modern Math - Permutation & Combination**

P . . . . . Q

R . . . . . S

T . . . . . U

V . . . . . W

Using 5 dots in each of the lines PQ, RS, TU and VW as the vertices, how many triangles can be drawn such that the base is on any one of the above lines?

- A.
120

- B.
150

- C.
200

- D.
600

Answer: Option D

**Explanation** :

The base has to be on one of the four lines.

Hence, the base can first be chosen in 4 ways.

Now, on this base, two points need to chosen from five to form the actual base. This can be done in ^{5}C_{2} ways i.e. 10 ways

For the third point, one of three lines has to be chosen and then one of five points from each line has to be chosen.

∴ Number of possible triangles = 4 × 10 × 3 × 5 = 600

Hence, option 4.

Workspace:

**IIFT 2016 QA | Modern Math - Permutation & Combination**

A playschool contains 4 boys and y girls. On every Wednesday during winter, five students, of which at least three are boys, go to Zoological Garden, a different group being sent every week. At the Zoological Garden, each boy in the group is given a ball. If the total number of balls distributed is 368, then the value of y is

- A.
5

- B.
6

- C.
7

- D.
8

Answer: Option D

**Explanation** :

Since the group has to have atleast three boys, number of ways in which a group of five can be formed = (^{4}C_{3} × * ^{y}*C

_{2}) + (

^{4}C

_{4}×

*C*

^{y}_{1}) = [4 × (

*y*)(

*y*– 1)/2] + [(1)(

*y*)] = 2

*y*(

*y*– 1) +

*y*

Since each boy in the group gets a ball, total balls distributed = 3[2*y*(*y* – 1)] + 4*y* = 6*y*^{2} – 2*y*

Since total balls distributed= 368; 6*y*^{2} – 2*y *= 368

∴ 3*y*^{2} – *y* – 184 = 0

On solving this, *y* = 8

Hence, option 4.

Workspace:

**IIFT 2016 QA | Modern Math - Permutation & Combination**

Which of the following statements regarding arrangement of the word ‘RIYADH’ is/are true:

i. Two vowels can be arranged together in 120 ways

ii. Vowels do not occur together in 240 ways

Which of the above statements are true?

- A.
Statement (i) only

- B.
Statement (ii) only

- C.
Both statements (i) and (ii)

- D.
None of the above

Answer: Option D

**Explanation** :

(i) Two vowels can be arranged together in RIYADH in 5! × 2! = 240 ways

Hence, statement (i) is not true.

(ii) Total arrangements where vowels are not together = total arrangements possible – arrangements where vowels are together = 6! – 240 = 720 – 240 = 480

Hence, statement (ii) is also not true.

Hence, option 4.

Workspace:

**IIFT 2016 QA | Modern Math - Permutation & Combination**

A reputed paint company plans to award prizes to its top three salespersons, with the highest prize going to the top salesperson, the next highest prize to the next salesperson and a smaller prize to the third-ranking salesperson. If the company has 15 salespersons, how many different arrangements of winners are possible (Assume there are no ties)?

- A.
1728

- B.
2730

- C.
3856

- D.
1320

Answer: Option B

**Explanation** :

This is a case of first selecting the top three salesmen out of 15 and then arranging them as first, second and third i.e. ^{15}P_{3}

∴ Number of arrangements = ^{15}C_{3} × 3! = 2730

Hence, option 2.

Workspace:

**IIFT 2016 QA | Modern Math - Permutation & Combination**

In an MBA entrance examination, a minimum is to be secured in each of the 6 sections to qualify the cut-offs. In how many ways can a candidate fail to secure the cut-offs?

- A.
60

- B.
61

- C.
62

- D.
63

Answer: Option D

**Explanation** :

Number of ways in which the candidate fails to secure the cut-offs in no section out of 6 sections = ^{6}C_{o }= 1

Number of ways in which the candidate may/may not secure the cut-offs

= ^{6}C_{o}+ ^{6}C_{1 } + ^{6}C_{2 } + ^{6}C_{3 }+ ^{6}C_{4 }+ ^{6}C_{5 }+ ^{6}C_{6}

= 2^{6} = 64

Thus, the number of ways in which he fails to secure the cut offs in at least one section = 64 − 1 = 63

Hence, option 4.

Workspace:

**IIFT 2015 QA | Modern Math - Permutation & Combination**

During the essay writing stage of MBA admission process in a reputed B-School, each group consists of 10 students. In one such group, two students are batchmates from the same IIT department. Assuming that the students are sitting in a row, the number of ways in which the students can sit so that the two batchmates are not sitting next to each other, is:

- A.
3540340

- B.
2874590

- C.
2903040

- D.
None of the above

Answer: Option C

**Explanation** :

Number of ways in which 10 students can sit = 10!

The number of ways in which two students (batchmates) sit together = 9! × 2

∴ The number of ways in which the student can sit so that the two batchmates are not sitting next to each other = 10! – 9! × 2 = 9! × 8 = 2903040

Hence, option 3.

Workspace:

**IIFT 2015 QA | Modern Math - Permutation & Combination**

In the board meeting of a FMCG Company, everybody present in the meeting shakes hand with everybody else. If the total number of handshakes is 78, the number of members who attended the board meeting is:

- A.
7

- B.
9

- C.
11

- D.
13

Answer: Option D

**Explanation** :

Let n members attended the board meeting.

Number of handshakes = n×(n – 1)/2 = 78

Solving this, n = 13

Hence, option 4.

Workspace:

**IIFT 2014 QA | Modern Math - Permutation & Combination**

In a school, students were called for the Flag Hoisting ceremony on August 15. After the ceremony, small boxes of sweets were distributed among the students. In each class, the student with roll no. 1 got one box of sweets, student with roll number 2 got 2 boxes of sweets, student with roll no. 3 got 3 boxes of sweets and so on. In class III, a total of 1200 boxes of sweets were distributed. By mistake one of the students of class III got double the sweets he was entitled to get. Identify the roll number of the student who got twice as many boxes of sweets as compared to his entitlement.

- A.
22

- B.
24

- C.
28

- D.
30

Answer: Option B

**Explanation** :

(1 + 2 + 3 + … + n) + x = 1200

Where, x is the roll number of the student who got twice as many as compared to his entitlement.

n (n + 1)/2 = 1200 – x ⇒ n (n + 1)

= 2400 – 2x

Substituting values of x from options, only for x = 24, the RHS can be expressed as a product of two natural numbers.

Hence, option 2.

Workspace:

**IIFT 2014 QA | Modern Math - Permutation & Combination**

In the MBA Programme of a B – School, there are two sections A and B. 1/4th of the students in Section A and 4/9th of the students in section B are girls. If two students are chosen at random, one each from section A and Section B as class representative, the probability that exactly one of the students chosen is a girl, is:

- A.
23/72

- B.
11/36

- C.
5/12

- D.
17/36

Answer: Option D

**Explanation** :

Selecting a girl from section A and section B is 1/4 and 4/9 respectively.

Selecting a boy from section A and section B is 3/4 and 5/9 respectively.

Case 1: A girl from section A and a boy from section B.

P_{1} = (1/4) × (5/9) = 5/36

Case 2: A boy from section A and a girl from section B.

P_{2} = (3/4) × (4/9) = 12/36

Required probability = P_{1} + P_{2} =17/36

Hence, option 4.

Workspace:

**IIFT 2014 QA | Modern Math - Permutation & Combination**

In an Engineering College in Pune, 8 males and 7 females have appeared for Student Cultural Committee selection process. 3 males and 4 females are to be selected. The total number of ways in which the committee can be formed, given that Mr. Raj is not to be included in the committee if Ms. Rani is selected, is:

- A.
1960

- B.
2840

- C.
1540

- D.
None of the above

Answer: Option C

**Explanation** :

Selecting 3 males from 8 and 4 females from 7 can be done in ^{8}C_{3} × ^{7}C_{4} = 56 × 35 = 1960

Raj and Rani together cannot be in the committee.

Selection of both Raj and Rani can happen in ^{7}C_{2} × ^{6}C_{3} = 21 × 20 = 420

Required number of ways = 1960 – 420 = 1540

Hence, option 3.

Workspace:

**IIFT 2013 QA | Modern Math - Permutation & Combination**

Mrs. Sonia buys Rs. 249.00 worth of candies for the children of a school. For each girl she gets a strawberry flavoured candy priced at Rs. 3.30 per candy; each boy receives a chocolate flavoured candy priced at Rs. 2.90 per candy. How many candies of each type did she buy?

- A.
21, 57

- B.
57, 21

- C.
37, 51

- D.
27, 51

Answer: Option B

**Explanation** :

Let the number of strawberry and chocolate flavoured candies be a and b respectively.

∴ 3.3a + 2.9b = 249

Since there is only one equation with two unknowns, substitute the given options into the equations.

By substitution, a = 57 and b = 21 satisfy the given equation.

Hence, option 2.

Workspace:

**IIFT 2013 QA | Modern Math - Permutation & Combination**

Out of 8 consonants and 5 vowels, how many words can be made, each containing 4 consonants and 3 vowels?

- A.
700

- B.
504000

- C.
3528000

- D.
7056000

Answer: Option C

**Explanation** :

Out of 8 consonants, 4 consonants can be selected in ^{8}C_{4 } = 70 ways.

Out of 5 vowels, 3 vowels can be selected in ^{5}C_{3}

= 10 ways.

These 7 selected letters can be arranged among themselves in 7! ways.

Thus total number of ways = 70 ´ 10 ´ 7! = 3528000

Hence, option 3.

Workspace:

**IIFT 2013 QA | Modern Math - Permutation & Combination**

In a sports meet for senior citizens organized by the Rotary Club in Kolkata, 9 married couples participated in Table Tennis mixed double event. The number of ways in which the mixed double team can be made, so that no husband and wife play in the same set, is

- A.
1512

- B.
1240

- C.
960

- D.
640

Answer: Option A

**Explanation** :

Two men can be selected in ^{9}C_{2} ways.

After selecting two men, two women can be selected in ^{7}C_{2} ways from (9 – 2 = 7) women, so that no husband and wife play in the same set.

Also, these selected 4 people can be grouped in 2 ways.

∴ The total number of mixed double teams

= ^{9}C_{2 }×^{7}C_{2 }× 2 = 1512

Hence, option 1.

Workspace:

**IIFT 2010 QA | Modern Math - Permutation & Combination**

In how many ways can four letters of the word ‘SERIES’ be arranged?

- A.
24

- B.
42

- C.
84

- D.
102

Answer: Option D

**Explanation** :

The words SERIES has one R, one I, two Es and two Ss.

Four letters can be selected and arranged in the following ways:

∴ Total number of arrangements = 6 + 12 + 12 + 12 + 12 + 12 + 12 + 24 = 102

Hence, option 4.

Workspace:

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