Probability and combinatorics questions

Question

If there are 20 people with 17 men and 3 women and they seat themselves randomly at 4 Tables (A,B,C,D) with 5 people each. With all arrangements being equally likely, what is the probability that no woman sits at table A ?

I am a little confused as how to being this problem. The total seating arrangements is 20 factorial but I am not sure how to handle duplicates here.

Answer 1

The first person to sit at A has a 17/20 chance of being a man, the second 16/19, third 15/18 and so on. Thus, the chance that only men are seated at table A is

(17*16*15*14*13)/(20*19*18*17*16)

Answer 2

I think you just want ₁₇C₅/₂₀C₅. What I have assumed is that Table A is filled first and the other tables after.

Then he question is just pick 5 men from 17 and no women from the whole group; the probability is calculated by dividing by the number of ways of picking 5 people from 20.

Please note: if I have misdirected you, please let me know when you review this problem.