Combination on jury selection $20$ women (including Alice and Betty) and $12$ men show up for jury duty. In how many ways can you select a jury of at least $5$ women and at least $5$ men if one of Alice or Betty must be selected, but they cannot both be selected? This is what I think: out of 12 men, choose 5, so that is a $_{12}C_5$, for women, you can only choose Betty or Alice, so there is $1$ choice there. If you have chosen one, the other cannot be chosen, so there are only $18$ women to choose from, and we need $4$ more women, so that is a $_{18}C_4$. Is it correct that my final answer would be $C(12,5)*1*C(18,4)$

$$\binom{18}{4}\binom{2}{1}\binom{12}{7}+\binom{18}{5}\binom{2}{1}\binom{12}{6}+\binom{18}{6}\binom{2}{1}\binom{12}{5}$$ The first term standing for a choice of $4$ women out of the women where Alice and Betty are excluded, of $1$ out of the couple Alice/Betty and $7$ out of all men.

The second term standing for a choice of $5$ women out of the women where Alice and Betty are excluded, of $1$ out of the couple Alice/Betty and $6$ out of all men.

The third term standing for a choice of $6$ women out of the women where Alice and Betty are excluded, of $1$ out of the couple Alice/Betty and $5$ out of all men.