From a group of 8 women and 6 men, a committee consisting of 3 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?

Answer:a) 2016ways b) 160ways c) 264ways

Step-by-step explanation:

a) Since selection are being made, we will use the combination rule.

If two men refuse to serve together, this means that only 1 man will be selected out of the three men that suppose to form a committee.

In order to select r objects from a pool of n objects, it is expressed as

nCr = n! / (n-r) ! r!

To select 3women from 8 women, we have 8C3

To select 3men from 6men with two men refusing to serve together we have 6C (3-2) i. e 6C1

The final selection can then be made in 8C3 * 6C1 ways

= 8! / (8-3) ! * 6! / (6-1) !1!

= 8!/5! * 6!/5!1!

= (8*7*6*5!/5!) * (6*5!/5!)

= 8*7*6 * 6

= 2016ways

b) Similarly if two of the women refuse to serve together, we will have;

8C1*6C3

That is only one woman is being selected out of 3.

8!/7!*6!/3!3!

= 8*6*5*4*3!/6*3!

= 8*5*4

= 160ways

c) If 1man and 1woman refuse to serve together, we will have selection of 2women and 2men from 8women and 6men to give us

8C2*6C2

= 8!/6!2!*6!/4!2!

= 8!/2!*1/4!2!

= 8*7*6*5*4!/4!*2

= 4*7*6*5

= 640ways