This is a combination problem. Order does not matter and the books are all different. You would write this as 21C5 or [tex] \tbinom{21}{5} [/tex]. Combinations are not fractions.
The answer is [tex] \frac{21 * 20 * 19 * 18 * 17}{1 * 2 * 3 * 4 * 5} [/tex]
which = 20349 different ways these books can be chosen.
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.
[tex]_{r}^{n}\textrm{C}=\frac{n!}{r!(n-r)!}[/tex]
Different ways to choose 5 books from 21 tiles:
[tex]_{5}^{21}\textrm{C}=\frac{21!}{5!(21-5)!}[/tex]
[tex]_{5}^{21}\textrm{C}=\frac{21!}{5!(16)!}[/tex]
[tex]_{5}^{21}\textrm{C}=\frac{21*20*19*18*17*16!}{5*4*3*2*1*(16)!}[/tex]
[tex]_{5}^{21}\textrm{C}=\frac{21*19*18*17}{3*2}[/tex]
[tex]_{5}^{21}\textrm{C}=\frac{21*19*3*17}{1}[/tex]
[tex]_{5}^{21}\textrm{C}=20,349[/tex]
Therefore, there are 20349 possible selections.
