Probability Seating A conference room has $m$ men and $w$ women and $m + w$ chairs.


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*Two of the men always sit together. Find the probability that all women are adjacent to each other.

*Find the probability that no two women are adjacent to each other
Please be detailed in your answer so I can understand clearly what you're saying.
 A: In the absence of information, I assume that seats are arranged in rows (or in a $\bigcap$ which amounts to the same thing).

Part 1
There are 3 groups: 2 men as one clump, $m-2$ individual men, and $w$ women as another clump. i.e. $(1+ m-2 +1) = m$ entities, and permute each clump among its members.
Thus 
$$\text{Favorable Ways} = (1+m-2+1)!\cdot2!\cdot w! = 2!\cdot m!\cdot w!$$
and
$$Pr = \dfrac{2!\cdot m!\cdot w!}{(m+w)!}$$

Part 2
$$\_m\_ m\_ m\_\ \dots\dots\ \_m\_ m\_ m\_$$
There are $m+1$ gaps between men (including ends) where women can be accommodated, and assuming that $w \le m+1$,
$$\text{Favorable Ways} = {m+1\choose w}\cdot w!\cdot m!$$
and
$$Pr = \dfrac{{m+1\choose w}\cdot w!\cdot m!}{(w+m)!}$$
A: Since it is a conference room, I am going to presume that people are seated in a circle, or that nobody has only one person next to them. 
So we can say that without any constraints, there are $(m + w - 1)!$ ways to place everyone, since we have to place one person, and then everyone else relative to them. If the table wasn't a circle, it would be the regular $(m + w)!$
If we go in one direction, say left to right, Then A sits next to his friend B  $\frac{1}{m + w - 1}$ or 1 out of all the possible people are the case we want. The same if we go from right to left, giving a final chance of $\frac{2}{m + w - 1}$
Another solution is to treat the two buddies as one, and sit them down, then we have $(m + w - 2)!$ ways to sit everyone else, and $2!$ ways of sitting the friends. So the number of cases where it happens out of all the possible cases is $$\frac{2!(m + w - 2)!}{(m + w - 1)!}$$
$$= \frac{2!(m + w - 2)!}{(m + w - 1)(m + w - 2)!} $$
$$= \frac{2}{m + w - 1} $$

This one is a bit trickier. We have to presume that m >= w
If we let all the guys sit down first, we have $(m-1)!$ ways of doing that, and they can leave $m$ possible gaps for the women to sit in. 
Now the problem is how many ways can the gaps be filled? It is $${{n} \choose {k} }= \frac{n!}{(n-k)!}$$
$${{m} \choose {w} }= \frac{m!}{(m-w)!}$$
Multiplying both parts together we get
$$\frac{m!(m - 1)!}{(m-w)!}$$ ways out of $(m + w - 1)!$
$$= \frac{\frac{m!(m - 1)!}{(m-w)!}}{(m + w - 1)!}$$
$$= \frac{m!(m-1)!}{(m-w)!(m + w -1)!}$$
