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Permutations & Combinations

  1. Mar 31, 2005 #1
    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?
    Thanks in advance.
     
  2. jcsd
  3. Mar 31, 2005 #2
    For the first one, use the fact that [tex]_{n} P _{k} = \frac{n!}{(n-k)!}[/tex]
     
  4. Mar 31, 2005 #3
    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.
     
  5. Mar 31, 2005 #4
    Try expressing 40320 as a factorial.
     
  6. Mar 31, 2005 #5
    I suppose it would be 8! = (10 - n)! then? Still don't know what to do...
     
  7. Mar 31, 2005 #6
    if 8! = (10-n)!
    n has to equal 2.
     
  8. Mar 31, 2005 #7
    Oh, I see now.. don't know why I didn't before. Thanks!
     
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