Question

A teacher has 10 different books. In how many ways can the teacher select 4 books to bring to class?(1 point)

Answer 1

The problem of selecting 4 books from 10 different books can be solved using the combination formula, which 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.

In this case, \( n = 10 \) and \( r = 4 \).

Applying the combination formula:

\[
\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}
\]

Now, we calculate \( 10! \) as \( 10 \times 9 \times 8 \times 7 \times 6! \). This allows us to cancel \( 6! \) in the numerator and the denominator:

\[
\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4!}
\]

Now, calculate \( 4! \):

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

Substituting this back into the equation gives:

\[
\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{24}
\]

Calculating the numerator:

\[
10 \times 9 = 90
\]
\[
90 \times 8 = 720
\]
\[
720 \times 7 = 5040
\]

Now we divide by 24:

\[
\frac{5040}{24} = 210
\]

Thus, the number of ways the teacher can select 4 books from 10 different books is \( \boxed{210} \).