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The gym has 11 different types of machines in the weight room. Geoff has time to use only 3 of them this afternoon. How many di...Asked by Lena
The gym has 11 different types of machines in the weight room. Geoff has time to use only 3 of them this afternoon. How many different combinations of machines can Geoff choose from to use? Show your work or explain how you got your answer
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Answered by
GPT-4o mini
To find the number of different combinations of 3 machines that Geoff can choose from 11 different types of machines, we can use the combination formula:
\[
C(n, r) = \frac{n!}{r!(n - r)!}
\]
where \( n \) is the total number of items to choose from (in this case, 11 machines), \( r \) is the number of items to choose (in this case, 3 machines), and \( ! \) denotes factorial, the product of all positive integers up to that number.
1. Here, \( n = 11 \) and \( r = 3 \).
2. Plugging these values into the combination formula gives:
\[
C(11, 3) = \frac{11!}{3!(11 - 3)!} = \frac{11!}{3! \times 8!}
\]
3. We can simplify this calculation:
\[
C(11, 3) = \frac{11 \times 10 \times 9}{3 \times 2 \times 1}
\]
4. Calculating the numerator:
\[
11 \times 10 = 110
\]
\[
110 \times 9 = 990
\]
5. Calculating the denominator:
\[
3 \times 2 \times 1 = 6
\]
6. Dividing the results:
\[
C(11, 3) = \frac{990}{6} = 165
\]
Thus, the total number of different combinations of machines that Geoff can choose from is \( \boxed{165} \).
\[
C(n, r) = \frac{n!}{r!(n - r)!}
\]
where \( n \) is the total number of items to choose from (in this case, 11 machines), \( r \) is the number of items to choose (in this case, 3 machines), and \( ! \) denotes factorial, the product of all positive integers up to that number.
1. Here, \( n = 11 \) and \( r = 3 \).
2. Plugging these values into the combination formula gives:
\[
C(11, 3) = \frac{11!}{3!(11 - 3)!} = \frac{11!}{3! \times 8!}
\]
3. We can simplify this calculation:
\[
C(11, 3) = \frac{11 \times 10 \times 9}{3 \times 2 \times 1}
\]
4. Calculating the numerator:
\[
11 \times 10 = 110
\]
\[
110 \times 9 = 990
\]
5. Calculating the denominator:
\[
3 \times 2 \times 1 = 6
\]
6. Dividing the results:
\[
C(11, 3) = \frac{990}{6} = 165
\]
Thus, the total number of different combinations of machines that Geoff can choose from is \( \boxed{165} \).
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