A textbook search committee is considering 20 books for possible adoption. The committee has decided to select 5 of the 20 for future consideration. In how many ways can it do so?

What is 20!/(20-5)!

Well, if the textbook search committee is anything like me, they'll want to have some fun while making their selection. So let's approach this math problem with a touch of humor!

The committee has 20 books to choose from, and they want to select 5 for future consideration. That's like trying to pick a favorite book out of a library – not an easy task!

To calculate the number of ways they can do this, we can use a combination formula. The formula goes like this:

nCr = n! / (r!(n-r)!)

Where n is the total number of books (20) and r is the number of books to be selected (5). The exclamation mark represents factorial, which means multiplying a number by all positive integers less than it down to 1.

Now, let's plug in the numbers and calculate this.

20C5 = 20! / (5!(20-5)!)

20! = 20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5!

The factorials look like a party of multiplying numbers, don't they? Anyway, let's calculate this without losing our sense of humor.

20C5 = (20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5!) / (5!(20-5)!)

Now, simplify it a bit:

20C5 = (20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5!) / (5! × 15!)

Look at that! The 5! in the numerator cancels with the 5! in the denominator, leaving us with:

20C5 = (20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6) / 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6

Finally, all the factors in the numerator and the denominator cancel out nicely, leaving us with the answer:

20C5 = 20 × 19 × 18 × 17 × 16 = 15,504

So, the textbook search committee can select 5 books out of 20 in 15,504 different ways! That's a lot of options for them to consider – may the best textbooks win!

To find the number of ways the committee can select 5 books out of 20 for future consideration, you can use the concept of combinations.

The formula for combinations is given by:

nCr = n! / (r!(n-r)!)

where n is the total number of items and r is the number of items to be selected.

In this case, we need to find the value of 20C5.

Using the formula:

20C5 = 20! / (5!(20-5)!)

= 20! / (5! * 15!)

= (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1)

= 15,504

Therefore, there are 15,504 ways the committee can select 5 books out of the 20 for future consideration.

To solve this problem, we need to find the number of ways to select 5 books from a pool of 20 books.

This can be done using the concept of combinations. The formula for combinations is given by the formula:

C(n, r) = n! / (r! * (n-r)!)

Where n is the total number of items and r is the number of items being selected.

In this case, we want to find C(20, 5), which can be calculated as follows:

C(20, 5) = 20! / (5! * (20-5)!)
= 20! / (5! * 15!)

Now, let's break down the calculation:

20! = 20 * 19 * 18 * 17 * 16 * 15!

5! = 5 * 4 * 3 * 2 * 1

15! = 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

By substituting these values, we can simplify the calculation:

C(20, 5) = (20 * 19 * 18 * 17 * 16 * 15!) / (5 * 4 * 3 * 2 * 1 * 15!)

Canceling out the common factors, we get:

C(20, 5) = (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1)

Now we can calculate this expression:

C(20, 5) = (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1)
= 20,160

Therefore, there are 20,160 different ways for the textbook search committee to select 5 books from the 20 available options.