Question

How many different ways are there to choose 5 books from a shelf of 24 books if the order in which they are selected doesn't matter?(1 point)
Responses

22
22

42,504
42,504

7 13
7 13

5,100,480
5,100,480
Question 2
How many ways are there if the order does matter?(1 point)
Responses

22
22

42,504
42,504

5,100,480
5,100,480

7 13

Answer 1

To answer both questions, we can use combinatorial formulas.

**Question 1: How many different ways are there to choose 5 books from a shelf of 24 books if the order in which they are selected doesn't matter?**

When the order does not matter, we use the combination formula:

\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]

where \(n\) is the total number of books and \(r\) is the number of books to choose. In this case, \(n = 24\) and \(r = 5\).

Calculate:

\[
\binom{24}{5} = \frac{24!}{5!(24-5)!} = \frac{24!}{5! \cdot 19!}
\]

This simplifies to:

\[
\binom{24}{5} = \frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1} = \frac{24 \times 23 \times 22 \times 21 \times 20}{120}
\]

Calculating the numerator:

\[
24 \times 23 = 552, \quad 552 \times 22 = 12144, \quad 12144 \times 21 = 255024, \quad 255024 \times 20 = 5100480
\]

Now divide:

\[
\frac{5100480}{120} = 42504
\]

So the answer to Question 1 is: **42,504**.

---

**Question 2: How many ways are there if the order does matter?**

When the order matters, we use the permutation formula:

\[
P(n, r) = \frac{n!}{(n-r)!}
\]

Again, we have \(n = 24\) and \(r = 5\).

Calculate:

\[
P(24, 5) = \frac{24!}{(24-5)!} = \frac{24!}{19!}
\]

This simplifies to:

\[
P(24, 5) = 24 \times 23 \times 22 \times 21 \times 20
\]

In essence, we already calculated this product when we worked through the combination earlier.

So,

\[
24 \times 23 \times 22 \times 21 \times 20 = 5100480
\]

Thus, the answer to Question 2 is: **5,100,480**.

---

To summarize:

- The answer to Question 1 is **42,504**.
- The answer to Question 2 is **5,100,480**.