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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?

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    $\begingroup$ Your answer seems to be correct $\endgroup$ – user655800 Oct 2 '19 at 1:19
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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??

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    $\begingroup$ Both your method and mine yield the same result. $\endgroup$ – user3347814 Oct 2 '19 at 2:57

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