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9 votes
9 votes

There are $4$ women $P, Q, R, S$ and $5$ men $V, W, X, Y, Z$ in a group. We are required to form pairs each consisting of one woman and one man. $P$ is not to be paired with $Z$, and $Y$ must necessarily be paired with someone. In how many ways can $4$ such pairs be formed?

  1. $74$
  2. $76$
  3. $78$
  4. $80$
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Migrated from GO Mechanical 4 years ago by Arjun

3 Answers

Best answer
12 votes
12 votes

Required number of ways $=$ Total number of possible ways $- n(P$ is paired with $Z) - n(Y$ is not paired$) + n(Y$ is not paired AND also $P$ and $Z$ are paired together$)$

Total number of ways $= 5\times 4\times 3\times 2 = 120$

$n(P$ is paired with $Z) = 4\times 3\times 2 = 24$

$n(Y$ is not paired$) = 4\times 3\times 2\times 1 = 24$

$n(Y$ is not paired AND also $P$ and $Z$ are paired together$) = 3\times 2\times 1 = 6$

So, required number of ways $= 120 - 24 -24 + 6 = 78.$

Correct Answer: $C$

selected by
4 votes
4 votes

Total ways possible (without any restriction) = 5 x 4 x 3 x 2 = 120

Ways when P is paired with Z = 4 x 3 x 2 = 24 (P & Z are out of the picture)

Ways when Y is not paired with anyone = 4 x 3 x 2 x 1 = 24 (Y is out of the picture)

(We also need to consider overlapping of above two cases)

Ways when both Y is not paired AND also P & Z are paired together = 3 x 2 x 1 = 6 (P, Z, and Y are out of the picture)

 

Total number of ways (with restrictions)  = 120 - 24 - 24 + 6  (Adding 6 as we have included it in both 24's)

                                                                         = 78

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We can also think as follows:

 Total number of ways with restrictions = Total ways possible (without any restriction) - Ways which break the rules

                                                                       = 120 - ( 24 + 24 - 6) (Inclusion-Exclusion Principle)

                                                                       = 78

3 votes
3 votes

Explanation:- Take different cases and possibilities that can occur while solving the questions. 

Note that in question it is given that a pair is defined as 1 Woman with 1 Man , P must not be paired with Z and also that Y must necessarily be paired with someone. 

So we'll take 4 possible cases viz :- 

Case1: when P is paired with Y so we can pair it in a unique way that's why it's alloted 1 as the probability and then we're checking for Q that how many options are left for Q to be paired so we'll write 4 (out of 5 because Y is already allocated to P). Similarly we check for R and S and write their possible numbers in front of them .. Now multiply these numbers and keep it in order to add them at the last..

Repeat the above step for rest of the three cases. 

Now we will simply add all the products together and arrive to the answer . 

 

Answer:

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