# Choosing a committee of $3$ members from $5$ men and $2$ women, with at least $1$ women. Two approaches give different answers.

Consider the number of ways to choose a committee of $$3$$ members from $$5$$ men and $$2$$ women where there must be at least $$1$$ woman.

First, we can choose a woman for the committee which can be done in $$2 \choose 1$$ ways and then we can choose the rest from the 6 people left in $$6 \choose 2$$ ways which gives us $$2 \choose 16 \choose 2$$ $$= 30$$ ways.

But if we divide it into cases: first if we consider the case where there is only $$1$$ woman and $$2$$ men in the committee, we get $$2 \choose 15 \choose 2$$ $$= 20$$ ways. Then we consider the case where there are $$2$$ women and $$1$$ man in the committee. Here we get $$2 \choose 25 \choose 1$$ $$= 5$$ ways. Summing up we get a total of $$25$$ ways.

But why are the processes giving us different answers? What am I missing?

• Your first method counts those committees with two women twice. – lulu Feb 12 '19 at 11:48

More specifically: choosing $$W_1$$ as the first woman and then $$M_1,W_2$$ as the other two results in the same committee as choosing $$W_2$$ as the first woman and then $$M_1,W_1$$ as the other two, but your first method counts them as two different solutions.