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

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1. CAT 2023 QA Slot 1 | Modern Math - Permutation & Combination CAT Question

The number of all natural numbers up to 1000 with non-repeating digits is:

(a)
648
(b)
585
(c)
738
(d)
504

2. CAT 2022 QA Slot 1 | Modern Math - Permutation & Combination CAT Question

The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is

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3. CAT 2022 QA Slot 2 | Modern Math - Permutation & Combination CAT Question

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

(a)
1200
(b)
1420
(c)
1440
(d)
1480

4. CAT 2022 QA Slot 3 | Modern Math - Permutation & Combination CAT Question

The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is

(a)
2222
(b)
2592
(c)
2442
(d)
3333

5. CAT 2021 QA Slot 1 | Modern Math - Permutation & Combination CAT Question

The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

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6. CAT 2021 QA Slot 2 | Modern Math - Permutation & Combination CAT Question

The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is

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7. CAT 2021 QA Slot 3 | Modern Math - Permutation & Combination CAT Question

A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

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8. CAT 2020 QA Slot 2 | Modern Math - Permutation & Combination CAT Question

How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

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9. CAT 2020 QA Slot 3 | Modern Math - Permutation & Combination CAT Question

How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?

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10. CAT 2019 QA Slot 1 | Modern Math - Permutation & Combination CAT Question

With rectangular axes of coordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y + 1) or to (x + 1, y), is

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11. CAT 2018 QA Slot 1 | Modern Math - Permutation & Combination CAT Question

How many numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

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12. CAT 2018 QA Slot 2 | Modern Math - Permutation & Combination CAT Question

In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is

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Answer: 1098

Video Explanation

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Text Explanation :

The number of matches played in a group of ‘n’ peope = $\frac{n (n - 1)}{2}$

The number of girl vs girl matches in junior level = 153.

Therefore, $\frac{n (n - 1)}{2}$ = 153 ⇒ n = 18

Therefore there are 18 girls at junior level.
⇒ There are 43 - 18 = 25 boys at junior level.

The number of boy vs boy matches in senior level = 276.
Therefore, $\frac{n (n - 1)}{2}$ = 276 ⇒ n = 24

Therefore there are 24 boys at senior level.
⇒ There are 51 - 24 = 27 girls at senior level.

Therefore, we have

Therefore, the number of boys vs girls matches = 25 × 18 + 24 × 27 = 1098

Hence, 1098.

13. CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination CAT Question

The number of solutions (x, y, z) to the equation x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12 is

(a)
101
(b)
99
(c)
87
(d)
105

14. CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination CAT Question

Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

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15. CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination CAT Question

In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

(a)
16
(b)
20
(c)
14
(d)
15

16. CAT 2017 QA Slot 2 | Modern Math - Permutation & Combination CAT Question

In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

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17. CAT 2017 QA Slot 2 | Modern Math - Permutation & Combination CAT Question

How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

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Answer: 50

Video Explanation

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Text Explanation :

For the 4 digit number to be divisible by 6, the sum of the digits of the 4 digit number has to be a multiple of 3 and the unit (right most) digit has to be an even number.

The combination of 4 digits that add upto a multiple of 3 are (0, 2, 3, 4), (0, 2, 4, 6) and (2, 3, 4, 6)

Case 1: Combination (0, 2, 3, 4)
For this combination the required number can start with 2, 3 or 4.
First digit 2: The last digit can be 0 or 4. The 2^nd and 3^rd digit from the left can be one 0/4 or 3. So a total of 4 possibilities.
First digit 4: Here too the number of possibilities is 4 as the 2^nd and 3^rd digit from the left can be one of 0/2 or 3.
First digit 3: Digits 0, 2 and 4 can be in any order as the 2^nd, 3^rd and 4^th digit from the left. So a total of 3! or 6 possibilities.
So number of possibilities with combination (0, 2, 3, 4) = 4 + 4 + 6 = 14

Case 2: Combination (0, 2, 4, 6)
The first digit can be anyone of 2, 4 or 6 i.e., 3 possibilities. The 2nd digit can be anyone from 0/2/4/6 but apart from the one used in the left most digit i.e., 3 possibilities. The 3^rd digit from the left will be any one from 0/2/4/6 but apart from the 2 digits used in the leftmost or 2^nd leftmost digit i.e., 2 possibilities. The extreme right digit will be the last digit after 3 out of these 4 digits are used in the first 3 digits from the left.
So total number of possibilities = 3 × 3 × 2 = 18

Case 3: Combination (2, 3, 4, 6)
The units digit has to be one amongst 2, 4 or 6 i.e., 3 possibilities. So first 3 digits from left will be 3 and two amongst 2/4/6 (i.e., except that digit which is not chosen as the right most digit) and can be arranged in 3! or 6 ways
So number of possibilities with combination (2, 3, 4, 6) is 6 × 3 = 18

So number of 4 digits numbers that can be formed using digits 0, 2, 3, 4 and 6 is 14 + 18 + 18 = 50

Hence, 50.

Directions for next 2 questions:

The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town.

18. CAT 2008 QA | Modern Math - Permutation & Combination CAT Question

Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is

(a)
60
(b)
75
(c)
45
(d)
90
(e)
72

19. CAT 2008 QA | Modern Math - Permutation & Combination CAT Question

Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is

(a)
1170
(b)
630
(c)
792
(d)
1200
(e)
936

20. CAT 2008 QA | Modern Math - Permutation & Combination CAT Question

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

(a)
499
(b)
500
(c)
375
(d)
376
(e)
501

21. CAT 2008 QA | Modern Math - Permutation & Combination CAT Question

What is the number of distinct terms in the expansion of (a + b + c)²⁰?

(a)
231
(b)
253
(c)
242
(d)
210
(e)
228

Answer: Option A

Video Explanation

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Text Explanation :

Consider (a + b + c)²⁰

Each term in the expansion of (a + b + c)²⁰ can be written as ²⁰C_r × a^x × b^y × c^z, where (x + y + z) = 20

Number of distinct terms in the expansion will be same as the number of distinct pairs of x, y and z.

∴ We have to divide 20 into three parts which can be done by using the distribution rule = ^n+r-1C_r-1

Where,

n is number of things to be distributed.

r is number of parts into which the things are to be distributed.

∴ To divide 20 into 3 parts we have,

^20+3-1C_3-1 = ²²C₂ = $\frac{22 \times 21}{2 \times 1}$ = 11 × 21 = 231

Alternatively,
This can be solved without using much knowledge of permutations and combinations as follows,

(a + b + c)¹ = a + b + c [i.e. 3 terms = (1 + 2) terms]

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac [i.e. 6 terms = (1 + 2 + 3) terms]

(a + b + c)³ = a³ + b³ + c³ + 6abc + 3ab² + 3ac² + 3a²b + 3bc² + 3a²c + 3b²c [i.e. 10 terms = (1 + 2 + 3 + 4) terms]

Similarly,

(a + b + c)ⁿ will have (1 + 2 + 3 + … + (n + 1)) terms

∴ (a + b + c)²⁰ will have (1 + 2 + 3 + … + 21) = 231 terms

Hence, option (a).

Answer the next 2 questions based on the information given below.

Let S be the set of all pairs (i, j) where 1 ≤ i < j ≤ n and n ≥ 4. Any two distinct members of S are called “friends” if they have one constituent of the pairs in common and “enemies” otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1, 2) and (2, 3) are also friends, but (1, 4) and (2, 3) are enemies.

22. CAT 2007 QA | Modern Math - Permutation & Combination CAT Question

For general n, how many enemies will each member of S have?

(a)
n – 3
(b)
$\frac{1}{2} (n^{2} - 3 n - 2)$
(c)
2n - 7
(d)
$\frac{1}{2} (n^{2} - 5 n + 6)$
(e)
$\frac{1}{2} (n^{2} - 7 n + 14)$

Answer: Option D

Video Explanation

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Text Explanation :

Enemies of every pair are the pairs formed with all numbers other than the two in the member itself.

∴ If there are n elements then each member has

$^{(n - 2)} C_{2} = \frac{(n - 2) (n - 3) (n - 4)!}{2 (n - 4)!} = \frac{1}{2} (n^{2} - 5 n + 6) enemies$

Hence, option (d).

23. CAT 2007 QA | Modern Math - Permutation & Combination CAT Question

For general n, consider any two members of S that are friends. How many other members of S will be common friends of both these members?

(a)
$\frac{1}{2} (n^{2} - 5 n + 8)$
(b)
$2 n - 6$
(c)
$\frac{1}{2} n (n - 3)$
(d)
n - 2
(e)
$\frac{1}{2} (n^{2} - 7 n + 16)$

24. CAT 2007 QA | Modern Math - Permutation & Combination CAT Question

In a tournament, there are n teams T₁ , T₂ ....., T_n with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common:

T₁ & T₂ , T₂ & T₃ ,......, T_{n − 1} & T_n , and T_n & T₁.

No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?

(a)
n(k – 1)
(b)
k(n – 1)
(c)
n(k – 2)
(d)
k(k – 2)
(e)
(n – 1)(k – 1)

25. CAT 2006 QA | Modern Math - Permutation & Combination CAT Question

There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?

(a)
144
(b)
180
(c)
192
(d)
360
(e)
716

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

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