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.