Question

From a group of 6 men and 5 women.

a. How many different committees can be formed if one of the men must be in the committee?

b. How many different committees can be formed if 2 of the women are enemies and refuse to serve on the committee together?

Answer 1

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a. So, 1 man must be always be in the committee. Lets take him into committee , that leaves us with 5 men and 5 women

I can take 5 men in 5C0 + 5C1 + 5C2 + ...5C5 ways = 2^5 = 32 ways, and another 5 women in 32 ways, which makes the number of different types of committee possible

as 1*32*32 = 120*120 = 1024 ways

b. If 2 women are enemies then , lets assume a and b woman are enemies

If a is in committee then b can't be. So, we are left with 4 women , and 5 men which can make 2^4*2^6 = 16*64= 1024 committee but the same committee can be formed when b is the committee and a isn't. Which means 1024+1024 = 2048 committees can be formed.

Answer is 2048 ways can such a committee be formed