# Permutations & Combinations

What would be the best way to solve for n if 10Pn = 90?
Also, how would you solve this problem:
In a student council election, there are 3 candidates for president, 3 for secretary, and 2 for treasurer. Each student may vote for at least one position. How many ways can a ballot be marked?

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For the first one, use the fact that $$_{n} P _{k} = \frac{n!}{(n-k)!}$$

I used that and multiplied both sides by (10 - n)!, then divided both sides by 90. Then I got 40320 = (10 - n)!. But that's where I got stuck.

Try expressing 40320 as a factorial.

I suppose it would be 8! = (10 - n)! then? Still don't know what to do...

if 8! = (10-n)!
n has to equal 2.

Oh, I see now.. don't know why I didn't before. Thanks!