A textbook search committee is considering 11 books for possible adoption. The committee has decided to select 3 of the 11 for future consideration. In how many ways can it do so?

Choosing 3 out of 11 is given by:

11C3
=11!/[(11-3)!3!]
=11*10*9/(1*2*3)
=165

To calculate the number of ways the committee can select 3 out of 11 books for future consideration, we can use the concept of combination.

The formula for combination is nCr = n! / r!(n-r)!, where n is the total number of items and r is the number of items being selected.

In this case, n = 11 (the total number of books) and r = 3 (the number of books the committee wants to select).

Using the combination formula, we can calculate the number of ways:

11C3 = 11! / 3!(11-3)!
= 11! / 3!8!
= (11 * 10 * 9) / (3 * 2 * 1)
= 165

Therefore, there are 165 different ways the committee can select 3 books out of the 11 for future consideration.

To calculate the number of ways to select 3 books out of the 11 options, we can use the formula for combinations, which is given by:

C(n, k) = n! / (k!(n-k)!)

Where:
- C(n, k) represents the number of ways to choose k items from a set of n items without regard for the order.
- n! represents the factorial of n, which is the product of all positive integers from 1 to n.

Applying this formula, we can calculate the number of ways to select 3 books from the 11 options:

C(11, 3) = 11! / (3!(11-3)!)
= 11! / (3!8!)
= (11 * 10 * 9) / (3 * 2 * 1)
= 165

Therefore, there are 165 possible ways for the committee to select 3 books out of the 11 options.