# Is there a typo in the problem about factorials?

#### davedave

Here is the question with a possible error.

(ex) When solving $$\displaystyle \frac{3n(n-2)!}{(n-3)!}=105$$ algebraically for n, which answer is eliminated because of the restrictions on factorials?

(a) -5
(b) -4
(c) -3
(d) -2

The answer key says the answer is (a). But, I think all the answers can be eliminated because they all give negative factorials when you plug them into the equation. Do you think there is a typo? Thanks.

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#### skipjack

Forum Staff
There isn't a typo. However, the question is poorly worded.

When solving, one uses (n - 2)!/(n - 3)! = n - 2. That's not valid unless n > 2, which eliminates all the answers. However...

It remains to solve 3n(n - 2) = 105, which is satisfied by n = 7 and n = -5, but n = -5 must be eliminated because of the requirement that n > 2.

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