# Combinations from two groups

I am stumped on a problem with two parts. First part I think I might have done correctly, Second part im lost

From a group of 4 women and 6 men, In how many ways a committee consisting of 2 women and 3 men can be formed.
a) without any restrictions
b) if a one particular woman is not eligible to be on the committee
a) This one I think I understood. I had to find the possible combinations of the men and women and multiply them right? So $^4C_2.^6C_3$
b) I'm totally stuck on this one...

• (a) is fine. For (b), if one particular woman is not to be on the committee, how many women and how many men can you choose from? – David Feb 6 '14 at 6:08
• @David well, 3 women and 6 men? so is it similar to part "A" but instead with one less woman? – Krimson Feb 6 '14 at 6:10
• That's how I see it. Just had a thought though: what is meant by a committee? Do the positions matter, e.g. is "Ann as president, Betty as secretary" counted the same as "Ann as secretary, Betty as president"? If it's only the people that count and not the positions, then your answer is correct. And what you just said for (b) is correct too. – David Feb 6 '14 at 6:12
• @David You are thinking way out of the box ! – lsp Feb 6 '14 at 6:14
• Well, it is an important point. But I should think that the simpler interpretation is what was intended. – David Feb 6 '14 at 6:15

## 1 Answer

For part $b$, if one particular woman is restricted or not eligible, then you have only $(4-1)=3$ women to choose from !

Now you have reduced the problem similar to that as part $a$, but with $3$ women and $6$ men.So isn't the answer just: $^3C_2.^6C_3$

• thanks! its amazing how I over complicate even the most simplest thing... – Krimson Feb 6 '14 at 6:14
• @Krimson Never search for something that is not mentioned in the question. It will only complicate things. – lsp Feb 6 '14 at 6:18