1

A club consisting of 6 men and 9 women will choose a committee of 4. In how many ways can the committee be chosen if it must contain at least 1 man?

I started out this problem like this:

men C(6,1)
=6

women C(9,3) =84

But then I realized that it was exactly the same as my problem before this one. The one before said it must contain exactly 1 man which I double checked my answer on here and it was correct. Now this question is asking for it to contain at least 1 man, so now I'm confused on how to solve it.

Shelby Trees
  • 19

3 Answers3

2

Hint.

Unrestricted choice of 4 people out of 15 minus choice of 4 out of 9 women.

Here we are exploiting the ease of calculating the complement.

Shailesh
  • 3,789
1

So you've calculated the number of ways for exactly one man. Now determine the combinations for two men, three men and four men, and you'll have it.

Number of ways at least 1 man= number of ways 1 man 3 women + number of ways 2 men 2 women + number of ways 3 men 1 women + number of ways 4 men 0 women.

MegaboofMD
  • 116
0

The total number of possible combinations for picking 4 people out of 15 is $\binom{15}{4} = 1365$

The number of combinations for picking 4 women out of 9 is $\binom{9}{4} = 126$

The number of combinations that includes at least 1 man = the total number of combinations - the number of combinations that only include women

Simpson17866
  • 522