CAT Previous Year Questions: Probability & Combinatorics (July 23) - CAT MCQ

Let the number of balloons each child received be 2a, 2b, 2c and 2d

2a + 2b + 2c + 2d = 20

a + b + c + d = 10

Each of them should get more than zero balloons.

Therefore, total number of ways = 

The number of 4-digit numbers possible using 1,1,2, and 4 is 4! / 2! ​= 12

Number of 1's, 2's and 4's in units digits will be in the ratio 2:1:1, i.e. 6 1's, 3 2's and 3 4's.

Sum = 6(1) + 3(2) + 3(4) = 24

Similarly, in tens digit, hundreds digit and thousands digit as well.

Therefore, sum = 24 + 24(10) + 24(100) + 24(1000) = 24(1111)

Mean = 24(1111) / 12 = 2222

The answer is option A.

Case 1: 4-digit numbers

Given digits - 0, 1, 2, 3, 4, 5

_, _, _, _

As the numbers should be greater than 2000, first digit can be 2, 3, 4 and 5.

For remaining digits, we need to arrange 3 digits from the remaining 5 digits, i.e. 5*4*3 = 60 ways

Total number of possible 4-digit numbers = 4*60 = 240

Case 2: 5-digit numbers

_, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2 = 600

Case 3: 6-digit numbers

_, _, _, _, _, _

First digit cannot be zero.

Therefore, total number of cases = 5*5*4*3*2*1 = 600

Total number of integers possible = 600 + 600 + 240 = 1440

The answer is option A.

This question is an application of the product rule in probability and combinatorics.

In the product rule, if two events A and B can occur in x and y ways, and for an event E, both events A and B need to take place, the number of ways that E can occur is xy. This can be expanded to 3 or more events as well.

Event 1: Distribution of balloons

Since each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them. 

So we are left with 15 - 4 x 3 = 15 - 12 = 3 balloons and 3 children. 

Now we need to distribute 3 identical balloons to 3 children. 

This can be done in ​ ways, where n = 3 and r = 3. 

So, number of ways = 

Event 2: Distribution of pencils

Since each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with 6 - 3 = 3 pencils.

We now need to distribute 3 identical pencils to 3 children.

This can be done in  ways, where n = 3 and r = 3.

So, number of ways = 

Event 3: Distribution of erasers

We need to distribute 3 identical erasers to 3 children.

This can be done in ​ ways, where n = 3 and r = 3.

So, number of ways = 

Applying the product rule, we get the total number of ways = 10 x 10 x 10 = 1000.

The possible arrangements are of the form 

35 _ Can be chosen in 6 ways.

35 _ _ We can choose 2 out of the remaining 6 in ⁶C₂ ​= 15ways. We remove 1 case where 7 and 8 are together to get 14 ways.

35 _ _ _We can choose 3 out of the remaining 6 in ⁶C₃ = 20ways. We remove 4 cases where 7 and 8 are together to get 16 ways.

35 _ _ _ _We can choose 4 out of the remaining 6 in ⁶C₄ ​ = 15ways. We remove 6 case where 7 and 8 are together to get 9 ways.

35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.

Thus, total number of cases = 6+14+16+9+2 = 47.

Alternatively,

The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.

Considering the cases, only 7 is selected.

We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.

Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have

Each of the numbers can either be selected or not selected and we have 4 numbers :

Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.

SImilarly, including only 8, we have 16 more possibilities.

Cases including neither 7 nor 8.

We must have 3 and 5 in the group but there must be no 7 and 8 in the group.

Hence we have 3 5 _ _ _ _.

For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6.

= 16 possibilities

But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.

Hence 16 - 1 = 15 cases.

Total = 16+15+16 = 47 possibilities

Let the numbers be of the form 100a + 10b + c, where a, b, and c represent single digits.

Then (100c + 10b + a)-(100a + 10b + c)=198

99c - 99a=198

c - a = 2.

Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number.

Thus, there can be 7 cases.

B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be 
7  x 10 = 70

The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.

The different possibilities include :

Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are :

4! / 3!​ = 4.

Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are :

4! / 2!⋅2!  ​= 6

Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are :

4! / 3! = 4

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :

4! / 2! = 12

Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

4! / 2! = 12

Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are :

4! / 2! = 12

A total of 12 + 12 + 12 + 4 + 6 + 4 = 50 possibilities.

The number of paths from (1, 1) to (8, 10) via (4, 6) = The number of paths from (1,1) to (4,6) * The number of paths from (4,6) to (8,10)

To calculate the number of paths from (1,1) to (4,6), 4-1 =3 steps in x-directions and 6-1=5 steps in y direction

Hence the number of paths from (1,1) to (4,6) = ⁽³⁺⁵⁾C₃​ = 56

To calculate the number of paths from (4,6) to (8,10), 8-4 =4 steps in x-directions and 10-6=4 steps in y direction

Hence the number of paths from (4,6) to (8,10) = ⁽⁴⁺⁴⁾C₄​ = 70

The number of paths from (1, 1) to (8, 10) via (4, 6) = 56*70=3920

Let 'ab' be the two digit number. Where b ≠ 0.

We will get number 'ba' after interchanging its digit. 

It is given that 10a+b > 3*(10b + a)

7a > 29b

If b = 1, then a = {5, 6, 7, 8, 9}

If b = 2, then a = {9}

If b = 3, then no value of 'a' is possible. Hence, we can say that there are a total of 6 such numbers.

In a tournament, there are 43 junior level and 51 senior level participants.

Let 'n' be the number of girls on junior level. It is given that the number of girl versus girl matches in junior level is 153.

⇒ ⁿC₂ = 153

⇒ n(n-1)/2 = 153

⇒ n(n-1) = 306

⇒ n²-n-306 = 0

⇒ (n+17)(n-18)=0

⇒ n = 18  (rejecting n=-17)

Therefore, number of boys on junior level = 43 - 18 = 25. 

Let 'm' be the number of boys on senior level. It is given that the number of boy versus boy matches in senior level is 276.

⇒ ^mC₂ = 276

⇒ m = 24

Therefore, number of girls on senior level = 51 - 24 = 27. 

Hence, the number of matches a boy plays against a girl  = 18 x 25 + 24 x 27 = 1098