From a group of 5 women and 7 men. (a) How many different committees consisting of 2 women and 3 men can be formed? (b) What if 2 of the men are feuding and refuse to serve on the committee together?

Answer :

Answer:

(a). 350

(b). 300

Step-by-step explanation:

We have been given that there is group of 5 women and 7 men.

(a). We are asked to find the number of committees consisting of 2 women and 3 men.

We know that we can choose 2 women from 5 women in [tex]5C2[/tex] ways. We can choose 3 men from 7 men in [tex]7C3[/tex] ways.

So we can choose 2 women and 3 men in [tex]5C2\times 7C3[/tex] ways.

[tex]\frac{5!}{2!(5-2)!}\times \frac{7!}{3!(7-3)!}[/tex]

[tex]\frac{5!}{2!(3)!}\times \frac{7!}{3!(4)!}[/tex]

[tex]\frac{5*4*3!}{2*1(3)!}\times \frac{7*6*5*4!}{3*2*1(4)!}[/tex]

[tex]5*2\times 7*5[/tex]

[tex]10\times 35[/tex]

[tex]350[/tex]

Therefore, 350 committees can be formed consisting of 2 women and 3 men.

(b). We are asked to find the number of committees consisting of 2 women and 3 men, if 2 of the men are feuding and refuse to serve on the committee together.

To solve this problem, we need to find number of committees consisting two men from 5 members and subtract it from total number.

[tex]5C2*5C1=\frac{5!}{2!(5-2)!}*\frac{5!}{1!(5-1)!}[/tex]  

[tex]5C2*5C1=\frac{5!}{2!(3)!}*\frac{5!}{1!(4)!}[/tex]  

[tex]5C2*5C1=\frac{5*4*3!}{2*1(3)!}*\frac{5*4!}{1*(4)!}[/tex]

[tex]5C2*5C1=5*2*5[/tex]  

[tex]5C2*5C1=50[/tex]

[tex]300-50=300[/tex]

Therefore, 300 committees can be formed consisting of 2 women and 3 men, if 2 of the men are feuding and refuse to serve on the committee together.

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