There are 20 people. 16 men and 4 women. How many possibilities are there for 12-people list (= every person has a numbered place) such that all women must be listed.
My thought:
We will "promise" spots first for the women. And there are, $\left( {\begin{array}{*{20}{c}} {12} \\ 4 \\
\end{array}} \right)$ possibilities to place them.
Afterward, there are $16$ men, but only $8$ places left to complete the list. So,
$\left( {\begin{array}{*{20}{c}} {16} \\ 8 \\ \end{array}} \right)$ possibilities to choose and place them.
All in all, we have $\left( {\begin{array}{*{20}{c}} {12} \\ 4 \\ \end{array}} \right)\left( {\begin{array}{*{20}{c}} {16} \\ 8 \\ \end{array}} \right)$ possibilities.
Is that right?
$\endgroup$2 Answers
$\begingroup$It's not quite right, as you say that the place in the list is numbered. You use combinations here, which do not take into account the order. This is the way that I would do it:
Step 1: Pick the 12 people to go into the list. All 4 women must go in, so we choose 8 men ($C(16,8)$ choices) to give our selection of 12 people.
Step 2: Order the 12 people in the list. This is a permutation of the set of 12 people, which gives $12!$ orderings.
Hence the number of orderings where all 4 women are in the list is given by $C(16,8) \cdot 12!$
$\endgroup$ 1 $\begingroup$No; you choose positions for the women but there are $4!$ ways to permute them. Similarly for the men, you choose them but you can still permute them. Your answer is correct if you multiply by $4! \times 8!$.
$\endgroup$ 2
