Combinations question involving committee and restrictions

Question

In how many ways can a committee of three women and four girls be chosen from seven women and six girls so that if the eldest woman is serving on the committee then the youngest girl is not?

The answer is apparently, 375.

This is what I did to get my answer, however it is wrong. I want to know why?

Working out:

(Girl in one group, whilst elder is not) + (Elder in one group, whilst girl isn't) + (Both not in groups). This equates to:

($\binom{6}{3}$ $*$ $\binom{6}{4}$) + ($\binom{7}{3}$ $*$ $\binom{5}{4}$) + ($\binom{6}{3}$ $*$ $\binom{5}{3}$)

which is 395

Answer 1

You have forgotten to remove a spot in the committee after placing in one of them:

$$\binom{6}{3} \binom{\color{red}{5}}{\color{red}{3}} + \binom{\color{red}{6}}{\color{red}{2}}\binom{5}{4} + \binom{6}{3}\binom{5}{\color{red}{4}}=375$$

Error in first term: The youngest girl is already chosen and now there is one less spot for the remaining 5 girls

Error in second term: The eldest woman is already chosen and now there is one less spot for the remaining 6 women

Error in third term: There are 4 spots for girls, now excluding that youngest

Answer 2

Here is another quick way to approach it -

Unrestricted number of ways to form the committee: $\displaystyle {7 \choose 3} {6 \choose 4} = 525$

Number of ways to form the committee with both eldest woman and youngest girl in the committee: $\displaystyle {6 \choose 2} {5 \choose 3} = 150$

So, desired number of ways to form the committee is $525 - 150 = 375$.