To determine how many different combinations of 3 machines Geoff can choose from 11 different types of machines, we use the concept of combinations. The formula for combinations is given by:
\[ C(n, r) = \frac{n!}{r!(n - r)!} \]
where:
- \( n \) is the total number of items (in this case, machines),
- \( r \) is the number of items to choose,
- \( ! \) denotes factorial, which is the product of all positive integers up to that number.
For this problem:
- \( n = 11 \) (the total number of machines),
- \( r = 3 \) (the number of machines Geoff wants to choose).
Substituting the values into the combinations formula:
\[ C(11, 3) = \frac{11!}{3!(11 - 3)!} = \frac{11!}{3! \cdot 8!} \]
Next, we simplify \( 11! \) as follows:
\[ 11! = 11 \times 10 \times 9 \times 8! \]
So we can simplify the combinations calculation:
\[ C(11, 3) = \frac{11 \times 10 \times 9 \times 8!}{3! \times 8!} \]
The \( 8! \) cancels out from the numerator and the denominator:
\[ C(11, 3) = \frac{11 \times 10 \times 9}{3!} \]
Now, we calculate \( 3! \):
\[ 3! = 3 \times 2 \times 1 = 6 \]
Substituting this back into our equation gives:
\[ C(11, 3) = \frac{11 \times 10 \times 9}{6} \]
Calculating the numerator:
\[ 11 \times 10 = 110 \] \[ 110 \times 9 = 990 \]
Now, we divide by 6:
\[ C(11, 3) = \frac{990}{6} = 165 \]
Therefore, the number of different combinations of machines that Geoff can choose from is:
\[ \boxed{165} \]
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