Permutation and Combination Questions for CAT with Answers PDF

The first possibility is that all the three males are sitting together.
In this arrangement ,all the females will then be sitting together opposite their husbands.
Number of ways = 3 x 2 = 6.
The 2nd possibility is that a male and a female are sitting alternately.
There are 2 such ways. 
The number of ways is as shown in the diagrams below.

Case i:
Person goes directly:
Number of ways = 2
If he returns either directly without taking the same path or through B, then number of ways = 1 + 3 x 4 = 13
Total number of ways = 2 x 13 = 26

Case ii:
Person goes through B:
Number of ways = 3 x 4 = 12
If he returns either directly or through B without taking the same path, then number of ways = 2 + 2 x 3 = 8
Total number of ways = 12 x 8 = 96
Hence, total number of ways in both cases = 26 + 96 = 122

An HOD can be selected in four ways. Further, an Associate Professor can be selected in three ways, then an Assistant Professor can be selected in 2 ways and then there is only one way for the selection of a Lecturer.
Number of ways of selecting the faculty members = 4 × 3 × 2 × 1 = 4! = 24

For selection of student coordinators, the number of ways is ²⁰C₁₀.
Therefore, total number of ways to select a panel including the student coordinators = 4! × ²⁰C₁₀

If the HOD of Mechanical department is promoted to be the principal, the number of possibilities for the new HOD is 4 and the number of possibilities for the new Associate Professor is 8.
So, the total number of possibilities of changes in this case is 4 × 8 = 32.
Similarly, for Electrical department, Number of cases = 5 × 9 = 45
For Electronics department, Number of cases = 3 × 12 = 36
For Civil department, Number of cases = 5 × 5 = 25
For Chemical department, Number of cases = 2 × 6 = 12
For Computer Science department, Number of cases = 1× 7 = 7
For Production department, Number of cases = 4 × 6 = 24
Therefore, Total number of changes possible = 32 + 45 + 36 + 25 + 12 + 7 + 24 = 181

First Case: All three cards are face cards; Number of possible cases = ¹²C₃
Second Case: First, a number has to be selected; this can be done in 9 ways. Now, three cards of this number have to be selected, which can be done in ⁴C₃ ways.
Therefore, number of cases = 9 × ⁴C₃
Therefore, total number of cases = ¹²C₃ + 9 × ⁴C₃

A triangle can be formed by joining 3 non-linear dots, so ¹¹C₃ triangles would have been formed if no dots were in a straight line. But 5 dots, which are in a straight line, instead of giving ⁵C₃ triangles, gives no triangle. Therefore, required number of triangles = ¹¹C₃ – ⁵C₃ = (11 x 10 x 9 x 8!/(3!8!)) – (5 x 4 x 3!/(3!2!)) = 165 – 10 = 155

Case i: All of same colour WWW, BBB, RRR
In this case, total number of necklaces = 3
Case ii: 2 of same colour and 1 of other colour.
2 of same colour and 1 unique will have 6 cases.
These can be arranged in 6 × 3 = 18
Case iii: All different coloured beads in a necklace
The arrangements WBR, WRB, BRW, BWR, RWB and RBW will be considered as only one arrangement.
In this case, total number of necklaces = 1
So, total number of necklaces = 3 + 18 + 1 = 22

The hosts can take the seats at the short ends in 2! ways.
Note that the number of ways in which mn different items can be divided equally into m groups, each containing n objects and where the order of the groups is not important is given by:
Number of ways 

12 guests can be divided into 2 groups of 6 each in (12!)/[(6!)² (2!)] ways, (m = 2, n = 6).
They can then be assigned to either side of the long table in 2! x (12!)/[(6!)² (2!)] = (12!)/(6!)² ways.
Next on each side, 6 people can sit in 6! ways.
Hence, total number of ways = 2! x (6!)² x (12!)/(6!)² = 2(12!)

Since the letters CAT appear in the same place as in the word EDUCATION, the numbers of vowels (E, U, I, O) and consonants (D, N) that we need to arrange are 4 and 2, respectively.
Since no vowels are supposed to come together, the vowels can take 1st, 3rd, 7th or 9th places and the remaining places can be filled by consonants.
So, the required number of ways = 4! x 2! = 48

p, q and r can take values 1, 2, 3, ..., 8.
We know that 1 + w + w2 = 0, w3 = w6 = 1, w² = w⁵ = w⁸ and w = w⁴ = w⁷.
Thus, a selection has to be made by choosing exactly one digit from each of the brackets: (3, 6), (2, 5, 8) and (1, 4, 7).
This can be done in 2 x 3 x 3 = 18 ways.