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?
Best Answer
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.
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: