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
\[ \boxed{120} \]