Q304) combinatorics?

Question

a department contains 4 men and 6 women.
In how many ways can a committee of 3 members be formed if it must have at least one person of each gender?

Answer 1

Consider committees with 1 man and 2 women.

There are 4 ways to choose the man.

There are 6 ways to choose the first woman and 5 ways to choose the second. (6)(5) = 30 ways to choose a pair of women but this method counts each pair of women twice. 30 / 2 = 15 ways to choose a pair of women (their order of being chosen does not matter).

So, (4)(15) = 60 ways to choose a committee with 1 man and 2 women.

Consider committees with 2 men and 1 woman.

There are 4 ways to choose the first man and 3 ways to choose the second man. (4)(3) = 12 ways to choose a pair of men but this method counts each ordering of a pair of men (Tom and Harry, Harry and Tom). Since order does not matter, 12 / 2 = 6 ways of choosing a pair of men. There are 6 ways to choose 1 woman.

So, (6)(6) = 36 ways to choose a committee of 2 men and 1 woman.

60 + 36 = 96 ways to form a committee of 3 members with at least 1 member of each gender, given 4 men and 6 women.


Use combinations n choose r: C(n, r) = n! / ((n - r)!(r!))
where n! = n(n - 1)(n - 2)(n - 3)**(3)(2)(1).

[C(4, 1)][C(6, 2] + [C(4, 2)][C(6, 1)]
= [4! / ((3!)(1!))][6! / ((4!)(2!))] + [4! / ((2!)(2!))][6! / ((5!)(1!))]
= [4][15] + [6][6]
= 60 + 36
= 96

Answer: There are 96 ways to form a committee of 3 members with at least one person from each gender, given 4 men and 6 women.

Answer 2

4*6*8 = 192