[Math] How many ways can a committtee of $2$ women and $3$ men be selected from a group of $5$ women and $7$ men if two of the men refuse to serve together

combinatorics

I am trying to learn permutations and combinations. Please help me to solve the second part of the question.

From a group of $5$ women and $7$ men, how many different committees of $2$ women and $3$ men can be formed? What if $2$ of the men are feuding and refuse to serve on the committee together?

Best Answer

I gather that you understand that the number of committees of two women and three men that can be selected from five women and seven men is $$\binom{5}{2}\binom{7}{3}$$ Below is a hint for the second part of the question.

Hint: Consider three groups of people: the five women, the five men who are not feuding, and the two men who are feuding. There are two ways to form the committee so that the two men who are feuding are not both selected:

  1. Neither of the feuding men is selected. Choose two of the five women and three of the five men who are not feuding.
  2. Exactly one of the feuding men is selected. Choose two of the five women, two of the five men who are not feuding, and one of the two men who are feuding.
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