thank you!! and two more:

In how many ways can a committee of three women and 4 girls be chosen from 7 women and 6 girls so that if the eldest woman is serving on the committee, then the youngest girl is not? answer = 375

In how many ways can 4 physics books and 3 maths books be arranged on a shelf if a selection is made from 6 different physics books and 5 different maths books? in how many of these arrangements are all the physics books next to each other? answer = 6C4 * 5C3 * 7! = 756000, 86400

The first one can be down in two approaches. One approach is to go by cases. The other is to consider the complement - we cannot have both the eldest women and youngest girl on the committee.

Here is the complement approach. Without any restrictions, we simply choose from \( \binom73\binom64=525 \) ways. If both the eldest women and youngest girl are selected, then we need only 2 more women out of 6, and 3 more girls out of 5. Which is done in \( \binom62 \binom53 =150 \) ways.

Which works out, since \(525-150=375\).

(You may want to figure out how the cases work to build more intuition. I just provide the answer for now: \( \binom63\binom54 + \binom62\binom54 + \binom63\binom53=375\) ways.)

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The first part is the classic 'select-and-then-arrange' scenario, where we first actually choose the books in question. Then we deal with the arrangement only after this is done.

Now for the second one, begin in the same way. We select 4 out of 6 physics books and 3 out of 5 maths books in \( \binom64\binom53\) ways.

The four physics books \(P_1\), \(P_2\), \(P_3\) and \(P_4\) then need to be

*grouped* together into something like \( \boxed{P_1P_2P_3P_4}\). However, since no specific ordering of the physics books was required; just that they

*are* grouped, we first count the number of groups - \(4!\) ways.

With the physics books grouped, the problem now boils down to just arranging the four "objects" \(M_1\), \(M_2\), \(M_3\) and \( \boxed{physics} \). There are four objects here, and hence this is also doable in \(4!\) ways.

Final answer: \(\binom64\binom534!4!=86400\) ways.

Huh, weird... that's also the number of seconds in a day!