Math Stats Questions?

Question

in a stats class of 8 women and 8 men, 2 groups of 8 students are formed by random selection. what is the probability that all of the women are in the first group and all of the men are in the second group. (hint: find the number of ways arranging WWWWWWWWMMMMMMMM

i have an idea on how to solve. You can use NPR
16P8, would that be right?
N!/ (n-r)!

Answer 1

Got an idea......make 8 cards with the W and 8 cards with the M. Shuffle them, then put one on the left and then one on the right until they are dealt out. Record the outcome for this 100 times. This should give you a reasonable statistical place to start.

Answer 2

Once you choose 8, the other group is determined.

So, choose 8 times without replacement, from 16, order doesn't matter:

The number of groups of 8 that can be formed:
16C8 = k

Of these, only one has all men/women, or is it 2?

So, I think that the probability is 1/k or 2/k.

Answer 3

for the first person to be a woman, the chances are 8/16. for the second person to be a woman, the chances are 7/15. continuing, you get

8/16 * 7/15 * 6/14. . .

= 8!/(16!/8!) = 2 * (8!) / (16!)