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From a group of 4 men and 5 women, how many committees of size 3 are possible with 2 men and 1 woman, but 2 of the 4 men are feuding and refuse to serve on the committee together?

The answer I was given is to first calculate the total number of ways to select the men and women without restrictions, which is 4C2 * 5C1 = 30, and then subtract the ways to group those feuding men together, which is 2C2 * 5C1 = 5, giving the correct final answer as 25.

I don't understand why it is 2C2. Shouldn't it be 3C1? My reasoning behind this is that I group the pair of troublesome adults as one entity, so the 4 men "become" 3 units, and from there on, I choose 1 of the units.

Why is it 2C2?

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You are subtracting the "impossible" committee, which you have counted in the original calculation of $\binom{4}{2}\binom{5}{1}$ even though you shouldn't have. What are the impossible committees? They are precisely the ones where the feuding men both serve. This fixes the men: the impossible committees have as their only two men the two feuding men, and then any one woman. Thus, you have $\binom22 = 1$ option for the men, and $\binom51 = 5$ options for the woman, for a total of $\binom22\binom51$ options.

My reasoning behind this is that I group the pair of troublesome adults as one entity

It appears that you want to treat the feuding men as one entity because then you are automatically only counting situations in which they are both on the committee. However, this fails to get the correct answer for two reasons:

  1. By counting both men as one entity, you cannot put this entity together with another man on the committee, because that would give a total of 3 men. However, you are treating this entity as a single man, so you are overcounting.
  2. In your method of counting, you are still counting all committees where two non-feuding men serve. However, in this part of the argument, you are only trying to count committees which are ruled out because of the feud, because you want to subtract them from your total.
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The $_2C_2$ reflects the only possible way of choosing non-allowed committees:

  • Choosing the two feuding men from the two feuding men.

All other ways of choosing two men contain at least one man from the "non-feuding faction".

Maybe another approach would be:

  • $(_4C_2 \color{blue}{- 1})\cdot _5C_1$ where the $\color{blue}{\mbox{one}}$ not permitted way of choosing the two feuding men is subtracted.
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