CAT Previous Year Questions: Set Theory, Permutation & Combination (July 22) - CAT MCQ

Let n, s, d and t be the number of students who likes none of the drinks, exactly one drink, exactly 2 drinks and all three drinks, respectively.

It is given, 

n + s + d + t = 100 ...... (1)

s + 2d + 3t = 73 + 80 + 52

s + 2d + 3t = 205 ...... (2)

(2)-(1), we get

d + 2t - n = 105

Maximum value t can take is 52, i.e. t = 52, d = 1 and n = 0

Minimum value t can take is 5, i.e. t = 5, d = 95 and n = 0 (This also satisfies equation (1))

Difference = 52 - 5 = 47

The answer is option A.

Now 23 students choose maths as one of their subject.

This means (MPC)+ (MC) + (PC)=23 where MPC denotes students who choose all the three subjects maths, physics and chemistry and so on.

So MC + PM =5 Similarly we have PC+ MP =7 

We have to find the smallest number of students choosing chemistry

For that in the first equation let PM=5 and MC=0. In the second equation this PC=2

Hence minimum number of students choosing chemistry will be (18+2)=20 Since 18 students chose all the three subjects.

From observing the data given, we find that it is a closed 3 set Venn diagram.
Let the three sports be F, T and C for Football, Tennis and Cricket respectively
n(F U T U C) = 256 , n(F) = 144, n(T) = 123, n(C) = 132, n(F ∩ T) = 58, n(C ∩ T) = 25, n(F ∩ C) = 63

We know that (AUBUC) = n(A) + n(B) +n(C) - n(A ∩ B) - n(B ∩C) - n(C ∩ A) + n(A ∩ B ∩ C)
So, 256 = 144 + 123 + 132 - 58 - 25 - 63 + n (F ∩ T ∩ C)
n (F ∩ T ∩ C) = 256 - 144 + 123 +132 - 146
n (F ∩ T ∩ C) = 256 - 253 = 3
Now, it is easy to calculate the number of students who only play tennis using a Venn diagram.
n (Students who play only Tennis) = 123 - (55 + 3 + 22) = 123 - 80
n (Students who play only Tennis) = 43 students

Let us draw a Three set Venn diagram from values given
Since its given that n(H) = n(E) and n(only P) = 0
a + b + 20 + 10 = c + d + 20 + 10
a + b = c + d
As N = 74, a + b + 20 + 10 + c + d = 74
2(a + b) = 44 => a + b = 22
n(H) = a + b + 20 + 10 = 22 + 20 +10 = 52 students

Given that out of 200 students, 105 like Pizza and 134 like Burger
In order to find the possible number of students who like only Burger, we need to find the range between the minimum and maximum possible number of students

Minimum possible number of Students who like only Burger:

For this, we must try to optimize the Venn diagram in such a way that it has maximum intersection
No of Students who like only Burger = 134 – 105= 29

Maximum possible number of Students who like only Burger:

For this, we must optimize the Venn diagram in such a way that it has minimum intersection
We know that total number of Students = 200
So, Minimum number of students who like both = 105 + 134 – 200 = 39 students
Students who like Burger = 134
Max. no of Students who like only Burger = 134 – 39 = 95
So range => 29 ≤ n ≤ 95
From the options, only (D) 93 lies within range

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 x 2 x 2 x 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

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 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 :

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

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

Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are :
 
Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are :

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

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

Given that in a tournament there are 43 junior level and 51 senior level participants, let us assume that there are n number of girls and m number of boys
The number of girl versus girl matches in junior level is 153
So ⁿC₂ = 153

n(n − 1) / 2 = 153
n(n - 1) = 306 = 18 × 17
There are 18 girls and 43 - 18 = 25 boys.

Supposing that m boys in the senior level, ^mC₂ = 276
m(m − 1) / 2 = 276
m(m-1) = 552 = 276 × 2 = 138 × 4 = 69 × 8 = 23 × 24
In senior level there are 24 boys and 27 girls
The number of matches a boy plays against a girl has to be found
⟹ 25 × 18 + 24 × 27
⟹ 450 + 648
⟹ 1098
The number of matches a boy plays against a girl is 1098.

We have to find the number of solutions (x, y, z) to the given equation.
x – y – z = 25, where x, y, and z are positive integers such that x ≤ 40, y ≤ 12, and z ≤ 12.
x – y – z = 25, so the maximum value y and z can take is 12 and x can take is 40.
From 40 we can subtract some values to get 25 i.e.15 can be subtracted,
If x = 40 then y + z = 15
y ≤ 12 so y can take all the corresponding values 12, 11, 10,......3 it is a set of 10 values.
z ≤ 12, so z can take corresponding other values to get with 15 when added with x
Similarly,
If x = 39, then y + z = 14 here y can take all the value from 12, 11,...till 2
So there are 11 values.
If x = 38, then y + z = 13 here y can take all values from 12, 11, 10,....till 1
So there are 12 values.
If x = 37 then y + z = 12 here y can take 11, 10, 9,.....,1 so there are 11 values
If x = 36 then y + z = 11 here y can take 10, 9,......,1 so there are 10 values
If x = 35 then y + z = 10 here y can take 9, 8,.......,1 so there are 9 values
If x = 34 then y + z = 9 here y can take 8,.......,1 so there are 8 values
If x = 33 then y + z = 8 here y can take 7,........,1 so there are 7values
If x = 32 then y + z = 7 here y can take 6,........,1 so there are 6 values
If x = 31 then y + z = 6 here y can take 5.........,1 so there are 5 values
If x = 30 then y + z = 5 here y can take 4,......,1 so there are 4 values
If x = 29 then y + z = 4 here y can take 3,...,1 so there are 3values
If x = 28 then y + z = 3 here y can take 2,1 so there are 2 values
If x = 27 then y + z = 2 here y can take 1 so there is 1 value
x cannot be less than 27 because it is given that y and z are positive integers so it has to be at least one cannot be less than one
So, 12 × 13 / 2 + 21 = 78 + 21 = 99.
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 99.