Question 198768
<font face="Garamond" size="+2">


7 ways to pick the first book, then 6 ways to pick the second book, and so on...


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 7!]


For 3 of 7:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(n, r) = \frac{n!}{(n - r)!}]


So you want:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(7, 3) = \frac{7!}{(7 - 3)!} = \frac{7!}{4!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = 7 \times 6 \times 5]


You get to do your own arithmetic.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>