# A problem with my reasoning in a problem about combinations

I was given the following problem to solve:

A committee of five students is to be chosen from six boys and five
girls. Find the number of ways in which the committee can be chosen,
if it includes at least one boy.

My method was $\binom{6}{1}\binom{10}{4}= 1260$, using the logic of choosing $1$ boy, then choosing the rest. This was wrong, as the answer was $\binom{11}{5}-\binom{5}{5}= 461$.
The correct answer’s logic was committee with no restrictions – committee with no girls.

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Your method has two errors. First you are selecting one boy from 6 and the other four from the whole set of 10 (girls and boys). Further you are only selecting one boy when you should select all combination of 1, 2, 3, 4, 5 boys. Instead circumventing this route, you select five people from 11 people and subtract the term with zero boys Thus

it is either: ${11\choose5} – {5\choose5}$ or

it is $${6\choose1}{5\choose4}+{6\choose2}{5\choose3}+{6\choose3}{5\choose2}+{6\choose4}{5\choose1}+{6\choose5}{5\choose0}$$

Thanks

Satish

Your idea is to choose one boy and then four others, which might include further boys. Nice idea, but unfortunately it doesn’t work: the reason why should be clear from the following choices.

1. Choose the boy $B_1$, then four more people $G_1,G_2,G_3,B_2$.
2. Choose the boy $B_2$, then four more people $G_1,G_2,G_3,B_1$.

These are two of the committees you have counted. . .

. . . BUT they are actually the same committee, so you should not have counted it twice.

Similarly, by following your method, a committee $B_1,B_2,B_3,G_1,G_2$ would be counted three times, and so on. This is why your method gives the wrong answer.