# How many different selections of 4 books can be made from a bookcase displaying 12 books?

Feb 2, 2016

$495$

#### Explanation:

color(white)("X")_(12)C_4 = (12!)/((12-4)!*4!) = (cancel(12)xx11xxcancel(10)^5xx9)/(cancel(4)xxcancel(3)xxcancel(2)xx1) = 495

For each of the $12$ choices that could be made for the first book
there would be $11$ choices that could be made for the second book
then $10$ choices for the third
and $9$ choices for the fourth.

So there would be $12 \times 11 \times 10 \times 9$ ways of selecting the books
but the same 4 books could have been selected in $4 \times 3 \times 2 \times 1$ different ways (using the same logic as above).

So the actual number of different selections would be:
$\textcolor{w h i t e}{\text{XXX}} \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$