Integer Set Product Calculation
For how many three-element sets (order doesn't matter) of positive integers $ \{a,b,c\} $ is it true that $ a\times b\times c=2310 $? (The positive integers $ a, $ $ b, $ and $ c $ are distinct.)
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
The prime factorization of $ 2310 $ is $ 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 $. Therefore, we have the equation \[ abc = 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11,\]where $ a, b, c $ must be distinct positive integers and order does not matter. There are $ 3 $ ways to assign each prime number on the right-hand side to one of the variables $ a, b, c, $ which gives $ 3^5 = 243 $ solutions for $ (a, b, c) $. However, three of these solutions have two $ 1 $s and one $ 2310, $ which contradicts the fact that $ a, b, c $ must be distinct. Because each prime factor appears only once, all other solutions have $ a, b, c $ distinct. Correcting for this, we get $ 243 - 3 = 240 $ ordered triples $ (a, b, c) $ where $ a, b, c $ are all distinct.
Finally, since order does not matter, we must divide by $ 3!, $ the number of ways to order $ a, b, c $. This gives the final answer, \[\frac{240}{3!} = \frac{240}{6} = \boxed{40}.\]
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