Question

(Medal) How many ways can a group of 12 be selected from 6 men and 9 women if there must be 6 women, and no more than 4 men on the jury?

Answer 1

Are there answer choices or free response?

Answer 2

Does this question makes sense....

How many ways can a group of 12 be selected from 6 men and 9 women if there must be 6 women, and no more than 4 men on the jury?

If there cannot be more than FOUR men on the jury, then there must be, at a minimum EIGHT women on the jury and not SIX.

I still am not sure how to solve the, but the six women criteria might be added to confuse us.

Answer 3

I'm pretty sure the 6 women thing is to confuse us. What I'm wondering is if we have to count for each individual male/female or just generalize it, as if they are interchangeable.

Answer 4

This is free-response ofc. When I said 'must have 6 women', I meant at least 6 women.
I just simplified the question to make it easier to read. This is a problem with permutations/combinations (Fundamental Principle of Counting).

Answer 5

To make up a jury of \(12\) people with at least \(6\) women, you can have four possible arrangements: (Note: I'm not referring to the total number of choices here)
(1) \(6\) women, \(6\) men
(2) \(7\) women, \(5\) men
(3) \(8\) women, \(4\) men
(4) \(9\) women, \(3\) men

Obviously, options (1) and (2) are off the table since you want no more than \(4\) men on the jury.

So, from the selection of \(9\) women and \(6\) men, you can then have
\[\binom 98\binom64+\binom 99\binom63\]total ways to select a jury.