I have tried to solve this problem, but I have no way to check if my solution is right.
Here is what I though:
Total number of groups that can be formed out of 23 people (9 men, 14 women) will be $A = 2^{23}-1$ (Excluding a group with 0 people).
Total number of groups that can be formed out of 9 men will be $B = 2^9-1$
Total number of groups that can be formed out of 14 men will be $C = (2^{14})-1$
Therefor, the number of groups that can be formed out of 9 men and 14 with a least one man and one woman will be $A – B – C = (2^{23}) – (2^9) – (2^{14}) + 1$
If this is not the correct answer, please, show the right solution. If it is correct, is there a better way to find out without have to post a question here?
Best Answer
It is the number of ways you can choose at least one man times the number of ways you can choose at least one woman:
$$(2^9 - 1)(2^{14}-1) = 8371713$$
I frankly think this is a "better" way than yours, which you request.
Why do you care about the number of groups that can be formed from $14$ men??