From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. In how many ways can the committee be formed if one man and one woman refuses to serve tog

Answer:

910

Step-by-step explanation:

From the given information;

Suppose that everyone served together;

Then, we have:

= C(8,3) × C(6,3)

= [tex]\dfrac{8!}{3!(8-3)!}\times \dfrac{6!}{3!(6-3)!}[/tex]

= [tex]\dfrac{8!}{3!(5)!}\times \dfrac{6!}{3!(3)!}[/tex]

= 1120

But since 1 man and 1 woman refuses to serve together;

Then:

The combination will be:

= C(7,2) × C(5,2)

= [tex]\dfrac{7!}{2!(7-2)!}\times \dfrac{5!}{2!(5-2)!}[/tex]

= [tex]\dfrac{7!}{2!(5)!}\times \dfrac{5!}{2!(3)!}[/tex]

= 210

∴

1120-210 = 910

Thus, the number of ways committee can be formed = 910

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