Probability Question A conference room has m men and w women

Probability Question:

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

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

b. Find the probability that no two women are adjacent to each other

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I did another similar problem as follows, and have been trying to work the above one similarly, but am having trouble with the above. For a reference example, the previous problem was:

Similar problem completed: A conference room contains m men and w women. These people seat at random in m + w seats arranged in a row. Find the probability that all the women will be adjacent. Solution: P(women together) = [(m+1)!w!] / (m+w)!

Please show details on the above problem so I can understand how you are combining. Thank you so much.

Solution

Thisn all condition is for round chair

a. the m men and w women sit randomly, except that two specific men out of the m men always sit together.
With the two men sitting together, you can treat this as m-1 men and w women, and m+w-1 chairs.
There are (m+w-1)! ways to arrange the groups.

If you want the women to sit together, then they all must sit at an end, or between two men. The m-1 male groups can be arranged in (m-1)! ways. there are m places you can put the women. Viewing the chairs left to right, you can put the women to the left of any of the m-1 groups. The women can be arranged in w! ways.

So there are (m-1)! x m x w! =m! w! ways to satisfy the conditions.
The probability is m! w! / (m+w-1)!

b. You now have (m+w)! possible arrangements. To arrange all the men, in m! ways, and all the women, in w! ways,

With m men, you have m+1 places to put the women. Just choose w of those for their placement. By putting just 0 or 1 women in between the men, then the probability is 0 that no two women are adjacent.