Question

A librarian chooses seven holiday books from a selection of ten to be displayed in the window of the library. In how many different ways can she choose the group of seven books?

Answer 1

To determine the number of ways the librarian can choose 7 holiday books from a selection of 10, we use the concept of combinations, since the order in which the books are chosen does not matter.

The formula for combinations is given by:

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

where \( n \) is the total number of items to choose from, \( r \) is the number of items to choose, and \( ! \) denotes factorial, which is the product of all positive integers up to that number.

In this case, \( n = 10 \) and \( r = 7 \). We want to compute \( \binom{10}{7} \).

First, we can simplify \( \binom{10}{7} \) using the property of combinations that states \( \binom{n}{r} = \binom{n}{n-r} \). Thus, we have:

\[
\binom{10}{7} = \binom{10}{3}
\]

Now, we calculate \( \binom{10}{3} \):

\[
\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!}
\]

Next, we can simplify this expression. The factorial \( 10! \) can be expanded as follows:

\[
10! = 10 \times 9 \times 8 \times 7!
\]

Now substituting this back into our combination formula, we get:

\[
\binom{10}{3} = \frac{10 \times 9 \times 8 \times 7!}{3! \cdot 7!}
\]

The \( 7! \) in the numerator and denominator cancel out:

\[
= \frac{10 \times 9 \times 8}{3!}
\]

Next, we compute \( 3! \):

\[
3! = 3 \times 2 \times 1 = 6
\]

Now substituting this value back in, we have:

\[
\binom{10}{3} = \frac{10 \times 9 \times 8}{6}
\]

Calculating the numerator:

\[
10 \times 9 = 90
\]
\[
90 \times 8 = 720
\]

Now, dividing by \( 6 \):

\[
\frac{720}{6} = 120
\]

Thus, the total number of ways the librarian can choose 7 books from 10 is

\[
\boxed{120}
\]