Answer:
Step-by-step explanation:
You must use a combination:
[tex]_nC_k=\dfrac{n!}{k!(n-k)!}[/tex]
We have n = 16, k = 3.
Substitute:
[tex]_{16}C_3=\dfrac{16!}{3!(16-3)!}=\dfrac{13!\cdot14\cdot15\cdot16}{2\cdot3\cdot13!}\qquad\text{cancel}\ 13!\\\\=\dfrac{14\cdot15\cdot16}{2\cdot3}=\dfrac{7\cdot5\cdot16}{1}=560[/tex]
The number of possible selections is 560.
The library is to be given 3 books as a gift. The books will be selected from a list of 16 titles.
Here we used the combination
[tex]= nC_n\\\\= 16C_3\\\\= \frac{16!}{3!(16-3)!}\\\\ = \frac{16!}{3!13!}\\\\ = \frac{16\times 15\times 14\times 13!}{13!3!}\\\\ = \frac{16\times 15\times 14}{3\times 2\times 1}\\[/tex]
= 560
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