Anonymous

# Probability Question (Combination/Permutations)?

From a group of 7 men and 9 women a committee consisting of 4 men and 3 women is to be formed. How many different committees are possible if

(a) 2 of the men refuse to serve together?

(b) 2 of the women refuse to serve together?

(c) 1 man and 1 woman refuse to serve together?

Relevance

I have done all the work except the final values ine ach case - may see some minor errors but do contact by mail if you have some clarifcations

(a) 2 of the men refuse to serve together?

You can build a committee without the two particular men (call them X and Y) included by selecting 4 men out of 5 men(=7-2) men and 3 women out of 9 women in (5C4)(9C3) ways -------------A

The above selection excludes X and Y altogether.

Now if one of the 2 men (X or Y) is included

Selection of X or Y out of 2 is possible in 2C1 ways)

Then we can select the rest of the members of the committee by selecting 3 men (= 4-1) out of remaining 5 men(=7-2) in 5C3 ways and 3 women in 9C3 ways

Hence the case of having one of the 2 men on the committee is

(2C1)(5C3)(9C3) ways -----------------------B

Hence total no of ways for the selection of such a committee where both A and B are not included is A+B

= (5C4)(9C3) + (2C1)(5C3)(9C3) -------Ans

(b) 2 of the women refuse to serve together?

On similar lines we get the no of ways to form committee of

4 men and 3 women where 2 particular women don wish to be together

If both are excluded we have (7C4)(7C3) -------C

And if one is included we have (2C1)(7C4)(7C2) ---D

Hence total no of ways is

(7C4)(7C3) + (2C1)(7C4)(7C2) -----Ans

(c) Say 1 man (X) and 1 woman (Y) refuse to serve together?

Similar lines we have to consider formation of committee where we don’t have 1 particular man and woman together but here we have to go differently (3 steps)

Case 1 - leave both X and Y altogether we have to select 4 men from 6 (= 7-1) men and 3 women from 8 (= 9-1) women in (6C4)(8C3) ways -----------------------E

Case 2 – leave the particular man X out but include Y necessarily – here 1 women Y is already selected

So we select 4 out of 6 (=7-1) men and 2 women (=3-1)out of 8 (=9-1)women

In (6C4)(8C2) ways ------------------F

Case 3 - leave the particular woman Y out but include the man X necessarily – here 1 man X is already selected

So we select 3 (= 4-1) men out of 6 (=7-1) men and 3 women out of 8(=9-1) women

This is possible in (6C3)(8C3) ways --------G

So we have E + F + G or (6C4)(8C3) + (6C4)(8C2) + (6C3)(8C3) -----Ans

edit : minor error in b) corrected