Permutation & Combination CAT Previous Year Questions with Answer PDF

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.

Given that AB, CD, EF, GH and JK be five diameters of a circle with center at O.
We have to find by how many ways three points can be chosen out of A, B, C, D, E, F, G, H, J, K and O so as to form a triangle.
We know that any three points lying on the circle are non collinear.
There are 11 points here we have to choose from any three points that are not collinear.
From these 10 points A, B, C, D, E, F, G, H, J, K we can select any three points on the circle such that they are non collinear. We can take O and then from remaining 10 we can take remaining two.
Since AOB , COD , GOH , EOF , JOK cannot form triangles, 5 can be subtracted.

⟹ 120 + 45 – 5 = 160 ways.

Let us assume Amal gets 1 + x pens and Bimal gets 2 + y pens and Kamal gets 3 + z pens
1 + x + 2 + y + 3 + z = 8
x + y + z = 2 (x, y, z being whole numbers) or
x + (1 + y) + (3 + z )
x + y + z = 8 – 3 = 5 (x, y, z being naturals numbers)
Take 5 ones 1 _ 1 _ 1 _ 1 _ 1 , + symbol should be added in any 2 slots. (There are 4 empty slots)
This is done in ⁴C₂ ways = 6 ways.
Alternative method is to list down the possibilities as done below.

We have to find the no. of four digit numbers which are divisible by 6 that are to be formed using the digits 0, 2, 3, 4, 6.
We should know that 0 does not occur in the left most position. If the numbers are to be divisible by 6 it should be a multiple of 2 and 3.
So it should be an even number and also should be a multiple of 3.
⟹ 2 + 3 + 4 + 6 + 0 = 15
Here we have two choices, drop something such that the number should be a multiple of 3 or we can drop a multiple of 3.
The following are the possibilities,
Drop 3 ⟹ 2 + 4 + 6 + 0 = 12 which is a multiple of 3.
Drop 6 ⟹ 2 + 3 + 4 + 0 = 9 which is a multiple of 3.
Drop 0 ⟹ 2 + 3 + 4 + 6 = 15 which is a multiple of 3.
We have to check the possible outcomes,
For 2 , 4 , 6 , 0 ⟹ 3 × 3 × 2 × 1 = 18 ways.
For 2 , 3 , 4 , 0 ⟹ 3 × 3 × 2 × 1 = 18 ways.
But out of these 18 outcomes there will be some possibilities where the last digit is 3.
4 such possibilities exist here which needs to be eliminated.
Therefore 18 – 4 = 14 ways.
For 2, 3, 4 and 6 if we have last digit as 3 ie. _ _ _ 3, we can rearrange 2, 4 and 6 in 3! ways
Hence 2, 3, 4 and 6 can be arranged in 4! - 3! = 24 – 6 = 18 ways.
Therefore, in total we can form 18 + 14 + 18 = 50 four digit numbers.