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

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

**CAT 2022 QA Slot 1 | Modern Math - Permutation & Combination**

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

Answer: 84

**Explanation** :

Let the number of balloons each child gets is 2a, 2b, 2c and 2d respectively.

a, b, c, d > 0

⇒ 2a + 2b + 2c + 2d = 20

⇒ a + b + c + d = 10

Number of combinations of a, b, c and d such that a, b, c and d > 0 = ^{(10-4)+4-1}C_{4-1} = ^{9}C_{3} = 84.

Hence, 84.

Workspace:

**CAT 2022 QA Slot 2 | Modern Math - Permutation & Combination**

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

Answer: Option C

**Explanation** :

**Case 1:** Number of 4-digit numbers

__ __ __ __

Here,

unit’s digit can be either 2, 3, 4 or 5 i.e., 4 ways,

ten’s digit can be chosen from remaining 5 numbers in 5 ways,

hundred’s digit can be chosen from remaining 4 numbers in 4 ways.

thousand’s digit can be chosen from remaining 3 numbers in 3 ways.

⇒ Total such 4-digit numbers = 4 × 5 × 4 × 3 = 240

**Case 2:** Number of 5-digit numbers

__ __ __ __ __

Here,

unit’s digit can be either 1, 2, 3, 4 or 5 i.e., 5 ways,

ten’s digit can be chosen from remaining 5 numbers in 5 ways,

hundred’s digit can be chosen from remaining 4 numbers in 4 ways.

thousand’s digit can be chosen from remaining 3 numbers in 3 ways.

ten thousand’s digit can be chosen from remaining 2 numbers in 2 ways.

⇒ Total such 5-digit numbers = 5 × 5 × 4 × 3 × 2 = 600

**Case 2:** Number of 6-digit numbers

__ __ __ __ __

Here,

unit’s digit can be either 1, 2, 3, 4 or 5 i.e., 5 ways,

ten’s digit can be chosen from remaining 5 numbers in 5 ways,

hundred’s digit can be chosen from remaining 4 numbers in 4 ways.

thousand’s digit can be chosen from remaining 3 numbers in 3 ways.

ten thousand’s digit can be chosen from remaining 1 number in 1 way.

One lakh’s digit

⇒ Total such 6-digit numbers = 5 × 5 × 4 × 3 × 2 × 1 = 600

∴ Total required numbers = 240 + 600 + 600 = 1440.

Hence, option (c).

Workspace:

**CAT 2022 QA Slot 3 | Modern Math - Permutation & Combination**

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

Answer: Option A

**Explanation** :

8. Let us consider the sum of unit’s digit of all such numbers.

__ __ __ __ x __

If 1 occupies the units digit, number of such numbers = 3! = 6

Sum of all the units digit of such numbers = 6 × 1 = 6

If 2 occupies the units digit, number of such numbers = 3!/2! = 3

Sum of all the units digit of such numbers = 3 × 2 = 6

If 4 occupies the units digit, number of such numbers = 3!/2! = 3

Sum of all the units digit of such numbers = 3 × 4 = 12

∴ Sum of unit’s digits of all possible numbers = 6 + 6 + 12 = 24

Similarly,

Sum of ten’s digits of all possible numbers = 24 × 10 = 240

Sum of hundred’s digits of all possible numbers = 24 × 100 = 2400

Sum of thousand’s digits of all possible numbers = 24 × 1000 = 24000

⇒ Sum of all such numbers = 24 + 240 + 2400 + 24000 = 26664

Also, number of such numbers = 4!/2! = 12

∴ Average of all such numbers = 2664/12 = 2222

Hence, option (a).

Workspace:

**CAT 2021 QA Slot 1 | Modern Math - Permutation & Combination**

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

Answer: 47

**Explanation** :

Out of the seven integers given, 3 and 5 should always be selected.

**Case 1**: Exactly 1 of 7 or 8 is selected.

Exactly one of 7 or 8 can be selected in 2 ways.

Now we have already selected three integers.

For each of the remaining 4 integers, it may be selected or it may not be selected.

∴ Number of ways of selection for remaining integers = 2 × 2 × 2 × 2 = 16

∴ Total number of ways = 2 × 16 = 32

**Case 2**: None of 7 and 8 is selected.

So far we have selected only two integer.

For each of the remaining 4 integers, it may be selected or it may not be selected.

∴ Number of ways of selection for remaining integers = 2 × 2 × 2 × 2 = 16

But this includes one way when no integer is selected.

∴ Number of ways of selecting at least one integer = 16 – 1 = 15

∴ Total number of ways = 32 + 15 = 47.

Hence, 47.

Workspace:

**CAT 2021 QA Slot 2 | Modern Math - Permutation & Combination**

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

Answer: 1000

**Explanation** :

We know that number of distributing n identical objects amongst r different groups = ^{n+r-1}C_{r-1}

Distributing balloons

Since each child gets at least 4 balloons, we will first give 4 balloons to each of the 3 children.

Now we have 3 identical balloons to be distributed to 3 children.

Number of ways of doing this = ^{3+3-1}C_{3-1} = 10 ways

Distributing pencils

Since each child gets at least 1 pencil, we will first give 1 pencil to each of the 3 children.

Now we have 3 identical pencils to be distributed to 3 children.

Number of ways of doing this = ^{3+3-1}C_{3-1} = 10 ways

Distributing erasers

We have 3 identical erasers to be distributed to 3 children.

Number of ways of doing this = ^{3+3-1}C_{3-1} = 10 ways

∴ Total number of ways = 10 × 10 × 10 = 1000

Hence, 1000.

Workspace:

**CAT 2021 QA Slot 3 | Modern Math - Permutation & Combination**

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

Answer: 50

**Explanation** :

**Case 1**: Number contains 2, 3, 1, 1

Number of such numbers = 4!/2! = 12

**Case 2**: Number contains 2, 3, 3, 1

Number of such numbers = 4!/2! = 12

**Case 3**: Number contains 2, 2, 3, 1

Number of such numbers = 4!/2! = 12

**Case 4**: Number contains 2, 2, 3, 3

Number of such numbers = 4!/(2!×2!) = 6

**Case 5**: Number contains 2, 2, 2, 3

Number of such numbers = 4!/3! = 4

**Case 6**: Number contains 2, 3, 3, 3

Number of such numbers = 4!/3! = 4

∴Total such numbers = 12 + 12 + 12 + 6 + 4 + 4 = 50.

Hence, 50.

Workspace:

**CAT 2020 QA Slot 2 | Modern Math - Permutation & Combination**

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

Answer: 315

**Explanation** :

Out of 4 digits we already have 7 and 3. Now we have to select 2 more digits.

**Case 1**: zero is not selected.

We can select 2 more digits out of 7 in ^{7}C_{2} = 21 ways.

These 4 digits can be arranged in 4! = 24 ways

∴ Total number of ways = 21 × 24 = 504 ways.

Now, in half of these numbers 7 would be before 3 and in other half 3 would be before 7.

∴ 7 comes before 3 in 504/2 = 252 ways.

**Case 2:** zero is selected.

We can select 1 more digit out of 7 in ^{7}C_{1} = 7 ways.

These 4 digits can be arranged in 3 × 3! = 18 ways

∴ Total number of ways = 7 × 18 = 126 ways.

Now, in half of these numbers 7 would be before 3 and in other half 3 would be before 7.

∴ 7 comes before 3 in 126/2 = 63 ways.

∴ Total numbers in which 7 comes before 3 = 252 + 63 = 315.

Hence, 315.

Workspace:

**CAT 2020 QA Slot 3 | Modern Math - Permutation & Combination**

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

Answer: 252

**Explanation** :

Here we have to find all 3-digit numbers which have at least one digit repeated.

This can also be calculated by subtracting all 3-digit number which do not have any repeated digits from total 3-digit numbers.

Total 3-digit numbers = 9 × 10 × 10 = 900

3-digit numbers without repetition,

∴ The number 3-digit numbers without repetition = 9 × 9 × 8 = 648

Therefore, the number of 3-digit numbers which have at least one digit repeated 900 – 648 = 252.

Hence, 252.

Workspace:

**CAT 2019 QA Slot 1 | Modern Math - Permutation & Combination**

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

Answer: 3920

**Explanation** :

We have to go from (1, 1) to (8, 10) via (4, 6)

Number of paths from (1,1) to (8,10) = [Number of paths from (1,1) to (4,6)] × [Number of paths from (4,6) to (8,10)]

Path from (1,1) to (4,6):

Number of horizontal displacements (∆x) = 4 − 1 = 3 units and

Number of vertical displacements (∆y) = 6 − 1 = 5 units.

Hence, a total of 8 units.

∴ Number of paths from (1,1) to (4,6) = ^{8}C_{3} × ^{5}C_{5} = 56.

Path from (4,6) to (8,10):

Number of horizontal displacements (∆x) = 8 − 4 = 4 units and

Number of vertical displacements (∆y) = 10 − 6 = 4 units.

Hence, a total of 8 units.

∴ Number of paths from (4,6) to (8,10) = ^{8}C_{4} × ^{4}C_{4} = 70.

Total required number of paths = 56 × 70 = 3920.

Hence, 3920.

Workspace:

**CAT 2018 QA Slot 1 | Modern Math - Permutation & Combination**

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?

Answer: 502

**Explanation** :

As the digits appear in ascending order, there is only one way in which selected group of integers can be arranged.

Thus for two digit number, we need to select two digits.

This can be done in ^{9}C_{2} = 36 ways

As ^{n}C_{m} =^{9}C_{(n−m)} ; ^{9}C_{2} = ^{9}C_{7} = Number of ways in which number can have 2 or 7 digits = ^{9}C_{2} = ^{9}C_{7} = 36 ways

Number of ways in which number can have 3 or 6 digits = ^{9}C_{3} = 84 ways

Number of ways in which number can have 4 or 5 digits = ^{9}C_{4} = 126 ways

Number of ways in which number can have 8 digits = ^{9}C_{8} = 8 ways

Number of ways in which number can have 9 digits = 1 ways

1 + 8 + 2(36) + 2(84) + 2(126) = 502 numbers have 2 or more distinct digits such that the digits appear in ascending order.

Answer: 502

Workspace:

**CAT 2018 QA Slot 2 | Modern Math - Permutation & Combination**

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

Answer: 1098

**Explanation** :

The number of matches played in a group of ‘n’ people

$=\frac{n(n-1)}{2}$

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

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

Therefore there are 18 girls at junior level.

The number of boy vs boy matches in senior level = 276.

Therefore, $\frac{\mathrm{n}(\mathrm{n}-1)}{2}=276\mathrm{or}\mathrm{n}=24$

Therefore there are 24 boys at senior level.

Therefore, we have

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

Answer: 1098

Workspace:

**CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination**

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

Answer: Option B

**Explanation** :

x = 25 + y + z. The possible values of x, y, z and the corresponding number of values of y, z are tabulated below (x, y, z are positive integers).

We see that 27 ≤ x ≤ 40.

The number of solutions is 1 + 2 + … + 12 + 11 + 10 = 78 + 21 = 99.

Hence, option 2.

Workspace:

**CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination**

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?

Answer: 160

**Explanation** :

There are 5 pairs of diametrically opposite points and the centre O.

If O is not selected, the number of triangles = ^{10}C_{3} = 120.

If O is selected, the other two points can be selected in 10(8)/2, i.e., 40 ways. The number of triangles is 160.

Alternately,

No. of ways of choosing 3 points out of given 11 points = ^{11}C_{3} = 165.

Out of these 165 ways, 5 of these would give us 3 points on the diameters mentioned, which will not form a triangle.

Hence, no. of triangles = 165 - 5 = 160.

Hence, 160.

Workspace:

**CAT 2017 QA Slot 1 | Modern Math - Permutation & Combination**

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

Answer: Option A

**Explanation** :

After giving one eraser to each of the 4 kids, there are 3 left.

They can split 2, 1 or 1, 1, 1. (No kid can get 4)

There are ^{4}P_{2} + ^{4}C_{3}, i.e., 16 ways of distributing the erasers.

Hence, option 1.

**Alternately,**

Let the number of erasers given to the 4 kids be w, x, y, z.

w + x + y + z = 7.

After giving at least 1 eraser to each, 3 erasers will be left.

The number of positive integral solutions is ^{3 + 4 - 1}C_{4 - 1}, i.e. 20. This includes (4,1,1,1); (1,4,1,1); (1,1,4,1); (1,1,1,4).

The required number = 20 – 4 = 16.

Workspace:

**CAT 2017 QA Slot 2 | Modern Math - Permutation & Combination**

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?

Answer: 6

**Explanation** :

Out of 8 pens, Amal, Bimal and Kamal get at least 1, 2 and 4 pens. So out of 8, pens 8 – (1 + 2 + 3) or remaining identical pens are to be distributed amongst 3 people. So the number of ways in which 2 identical pens can be distributed amongst 3 people

= ^{2+3–1}C_{3–1 }= ^{4}C_{2} or 6 ways.

Answer: 6

Workspace:

**CAT 2017 QA Slot 2 | Modern Math - Permutation & Combination**

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?

Answer: 50

**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)

(i) 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 2nd and 3rd 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 2nd and 3rd 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 2nd, 3rd and 4th 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

(ii) 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 3rd digit from the left will be any one from 0/2/4/6 but apart from the 2 digits used in the leftmost or 2nd 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

(iii) 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

Answer: 50

Workspace:

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.

**CAT 2008 QA | Modern Math - Permutation & Combination**

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

Answer: Option D

**Explanation** :

We can find the number of shortest possible paths from A to E either by trial and error or by using combinations.

Note that to travel from A to E, we have to take 2 roads to the right and 2 roads downwards (in the diagram) in order that we follow the shortest path. In other words, we have to use 2 + 2 = 4 roads, out of which 2 are towards right and 2 are downwards.

This is equivalent to selecting 2 things (roads towards right) out of 4 things (roads). (The remaining two roads will be downwards.)

The number of ways of doing this is ^{4}C_{2} = 4!/(2!×2!) = 6

∴ From point A to E, there are 6 ways to reach with the minimum distance travelled.

Here E to F is the shortest distance because the third side of a triangle is always less than the sum of the other two sides.

From point F to B, there are ^{6}C_{4} = 6!/(4!×2!) = 15 ways to reach with the minimum distance travelled.

∴ There are 15 × 6 = 90 shortest paths that Neelam can choose.

Hence, option 4.

Workspace:

**CAT 2008 QA | Modern Math - Permutation & Combination**

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

Answer: Option A

**Explanation** :

From point A to B, there are 90 paths possible with the minimum distance travelled.

We can travel from B to C in two ways

1. B – K – M – C

To travel from B to K, we have to take 6 roads, out of which one is towards left and 5 are upwards.

There are 6!/(5!×1!) = 6 ways to do this. (Refer to the explanation in the first question of this set.)

2. B – N – C

To travel from B to N, we have to take 7 roads, out of which one is towards left and 6 are upwards.

There are 7!/(6!×1!) = 7 ways to do this.

∴ There are 6 + 7 = 13 ways in which we can travel from B to C.

∴ Overall there are 90 × 13 = 1170 paths possible

Hence, option (a).

Workspace:

**CAT 2008 QA | Modern Math - Permutation & Combination**

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

Answer: Option D

**Explanation** :

The minimum number that can be formed is 1000 and the maximum number that can be formed is 4000.

As 4000 is the only number in which the first digit is 4, first let us calculate the numbers less than 4000 and then we will add 1 to it.

∴ First digit can be 1, 2 or 3.

Remaining 3 digits can be any of the 5 digits.

∴ Total numbers that can be formed, which are less than 4000 = 3 × 5 × 5 × 5 = 375

∴ Total numbers that satisfy the given condition = 375 + 1 = 376

Hence, option (d).

Workspace:

**CAT 2008 QA | Modern Math - Permutation & Combination**

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

**Explanation** :

Consider (a + b + c)^{20}

∵ The degree of the expression is 20, the degree of each term of the expression after expansion will be 20.

∴ We have to divide 20 into three parts which can be done by using the distribution rule

^{n+r-1}C_{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+3-1}C_{3-1} = ^{22}C_{2} = $\frac{22\times 21}{2\times 1}$ = 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} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac [i.e. 6 terms = (1 + 2 + 3) terms]

(a + b + c)^{3} = a^{3} + b^{3} + c^{3} + 6abc + 3ab^{2} + 3ac^{2} + 3a^{2}b + 3bc^{2} + 3a^{2}c + 3b^{2}c [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).

Workspace:

**Answer the folowing question 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.

**CAT 2007 QA | Modern Math - Permutation & Combination**

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

- A.
n – 3

- B.
$\frac{1}{2}({n}^{2}-3n-2)$

- C.
2n - 7

- D.
$\frac{1}{2}({n}^{2}-5n+6)$

- E.
$\frac{1}{2}({n}^{2}-7n+14)$

Answer: Option D

**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

${}^{(n-2)}{C}_{2}=\frac{(n-2)(n-3)(n-4)!}{2(n-4)!}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}({n}^{2}-5n+6)\mathrm{enemies}$

Hence, option (d).

Workspace:

**CAT 2007 QA | Modern Math - Permutation & Combination**

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.
$\frac{1}{2}({n}^{2}-5n+8)$

- B.
$2n-6$

- C.
$\frac{1}{2}n(n-3)$

- D.
n - 2

- E.
$\frac{1}{2}({n}^{2}-7n+16)$

Answer: Option D

**Explanation** :

Two members are friends if they have one element in common.

∴ All the members having one constituent as the common element are common friends.

There are (n – 3) such friends.

Also, one pair formed by the uncommon constituents of the two friends is a common friend.

∴ There are n – 3 + 1 = n – 2 common friends.

Hence, option (d).

Workspace:

**CAT 2007 QA | Modern Math - Permutation & Combination**

In a tournament, there are n teams T_{1} , T_{2} ....., T_{n} with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common:

T_{1} & T_{2} , T_{2} & T_{3} ,......, T_{n − 1} & T_{n} , and T_{n} & T_{1}.

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)

Answer: Option A

**Explanation** :

Each team has (k – 2) players to itself and shares 2 players with other two teams.

n pairs of teams have 1 player in common and there are n teams.

Total number of players = n(k – 2) + n

= nk – 2n + n

= nk – n

= n(k – 1)

Hence, option (a).

Workspace:

**CAT 2006 QA | Modern Math - Permutation & Combination**

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

Answer: Option A

**Explanation** :

Task 2 can be assigned in 2 ways (either to person 3 or person 4).

Task 1 can then be assigned in 3 ways (persons 3 or 4, 5 and 6)

The remaining 4 tasks can be assigned to the remaining 4 persons in 4! = 24 ways

∴ The assignment can be done in 24 × 2 × 3 = 144 ways

Hence, option 1.

Workspace:

**CAT 2005 QA | Modern Math - Permutation & Combination**

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

- A.
200

- B.
216

- C.
235

- D.
256

Answer: Option A

**Explanation** :

Let there be g girls and b boys.

Number of games between two girls = ^{g}C_{2}

Number of games between two boys = ^{b}C_{2}

∴ g(g – 1)/2 = 45

∴ g^{2} – g – 90 = 0

∴ (g – 10)(g + 9) = 0

Since the number of girls cannot be negative, ∴ g = 10

Also,

b(b – 1)/2 = 190

∴ b2 – b – 380 = 0

∴ (b + 19)(b – 20) = 0

∴ b = 20

∴ Total number of games = ^{(g + b)}C_{2} = ^{30}C_{2} = 30 × 29/2 = 435

∴ Number of games in which one player is a boy and the other is a girl = 435 – 45 – 190 = 200

Hence, option 1.

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