Question

A student has selected 8 books that she likes, but she has money only for 3 books. how many possible selections does she have?

Answer 1

Since the order of the books chosen doesn't matter, we use the combination formula to find the number of possible selections. Recall that a combination of k items from a set of n items is given by   $nCk= \frac{n!}{k!(n-k)!}$. 

In this problem, n = 8 and k = 3. So the number of possible selections is 
$8C3= \frac{8!}{3!(8-3)!}=56$. Thus the answer is 56.

Answer 2

The student has 6,720 selections to choose from. Because order doesn’t matter, the problem is a combinations function. It is solved by dividing the factorial of 8 by the factorial of 3. So (8*7*6*5*4*3*2*1)/(3*2*1)