Question

A textbook search committee is considering
13 books for possible adoption. The committee has decided to select 3 of the 13 for further consideration. In how many ways can it do​ so?

Answer 1

Number of ways to select 3 books from 13 books for adoption is 286 .

Step-by-step explanation:

A Permutation is an ordered Combination. When the order does matter it is a Permutation. There are basically two types of permutation:

• Repetition is Allowed: such as the lock above. It could be "333".

• No Repetition: for example the first three people in a running race. You can't be first and second.

Formula is given by:

$nC_{r} = \frac{n!}{r!(n-r)!}$

, where n is the number of things to choose from,  and we choose r of them,  no repetitions,  order matters. Here , n=13 , r=3.

⇒  $13C_{3} = \frac{13!}{3!(13-3)!}$

⇒  $13C_{3} = \frac{13!}{3!(10)!}$

⇒  $13C_{3} = \frac{13(12)(11)10!}{3(2)(10)!}$

⇒  $13C_{3} = 13(2)(11)$

⇒  $13C_{3} = 286$

∴ Number of ways to select 3 books from 13 books for adoption is 286 .