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)

Krish R.

asked • 12/15/18If 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.

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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)

I think you just want _{17}C_{5}/_{20}C_{5}. 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.

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Krish R.

What if you cannot assume the order of filling ?12/16/18