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I've got a problem where I have $m$-men and $w$-women in a line, and I'm trying to find the probability that all the women are adjacent. The answer I got is: $\frac{(m+1)!w!}{(m+w)!}$

The idea is to treat the women as a single unit. So in that case there are $(m+1)!$ ways to organize everyone, and then $w!$ ways to organize the women while they're adjacent. Thus the $(m+1)!w!$. The denominator is the total number of people factorial: $(w+m)!$

Can someone confirm if I've done this right, or point out where it's wrong? The math seems sound, but something about the $+1$ and the actual numerical answers when using example numbers makes me feel like I've made a mistake. I don't know if that's just bad probability intuition or not. Any help appreciated.

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    What do you mean by 'women are adjacent'. All women are adjacent? At least two women are adjacent? – 2017-01-31
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    Yes, all women are adjacent. Will update the original post. – 2017-01-31
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    It seems totally fine to me! – 2017-01-31

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Confirmed!

There are $(m+1)!~w!$ ways to arrange the men and women in a row with the women all collated; as you reasoned.   That is the size of the favoured event of: no men between any two women (is that what you meant?).

There are $(m+w)!$ ways to arrange the men and women in a row with no restriction.   That is the size of the sample space.

All atomic outcomes in this space have equal probability, so the probability of the favoured event is: $$\dfrac{(m+1)!~w!}{(m+w)!}$$

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    Thanks, I appreciate it. Probability intuition is just terrible. – 2017-01-31