So I've got this problem:

From a group of 10 people (4 women and 6 men) 5 will be chosen randomly to occupy the following places in a committee:

1 president
1 vicepresident
3 council representatives

1. If 3 men were chosen to occupy the last 3 places in the committee, how many different councils could there be?
A.4 B.6 C.15 D.20

2. After choosing the 5 people, an observer realizes that, of the first 4 students chosen, 3 are female and 1 is male. The observer ascertains that the fifth student chosen will have:
A. 2 times the probability of being male instead of female
B. 2 times the probability of being female instead of male
C. 3 times the probability of being male instead of female
D. 3 times the probability of being female instead of male

3. The probability of choosing 2 men and 3 women is equal to the probability of choosing:
A. 4 men and 1 woman
B. 1 man and 4 women
C. 3 men, 2 women
D. 5 men and no women

I only know for sure the answer to the first question, by using combinations as follows:

6! / 3!(6-3)! = 6! / 3!3! = 20, thus D

But I can't quite figure out what to do about the other 2. I've got an exam for university this weekend and I'm pretty stressed out. Can you please help me?
Thanks in advance.