Question

The library is to be given 5 books as a gift. the books will be selected from a list of 21 titles. if each book selected must have a different title, how many possible selections are there?

Answer 1

This is a combination problem. Order does not matter and the books are all different. You would write this as 21C5 or $\tbinom{21}{5}$. Combinations are not fractions.

The answer is $\frac{21 * 20 * 19 * 18 * 17}{1 * 2 * 3 * 4 * 5}$

which = 20349 different ways these books can be chosen.

Answer 2

Answer:

20,349.

Step-by-step explanation:

We have been that a library is to be given 5 books as a gift. the books will be selected from a list of 21 titles.

We will combinations to solve our given problem.

$_{r}^{n}\textrm{C}=\frac{n!}{r!(n-r)!}$

Different ways to choose 5 books from 21 tiles:

$_{5}^{21}\textrm{C}=\frac{21!}{5!(21-5)!}$

$_{5}^{21}\textrm{C}=\frac{21!}{5!(16)!}$

$_{5}^{21}\textrm{C}=\frac{21*20*19*18*17*16!}{5*4*3*2*1*(16)!}$

$_{5}^{21}\textrm{C}=\frac{21*19*18*17}{3*2}$

$_{5}^{21}\textrm{C}=\frac{21*19*3*17}{1}$

$_{5}^{21}\textrm{C}=20,349$

Therefore, there are 20349 possible selections.