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 \( \boxed{210} \).
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