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Question from Meghan, a student:

Hello,

I'm working on the beginning of the combinatorics unit in Math 12 in BC.
We just looked at using the nPr or n!/(n-r)! formula.

My questions for homework that I didn't quite understand how to do were:

Solve each equation for n.
1) nP2 = 90
2) 6Pn = 720

I managed to do these in my head, but I'm not sure how to solve these algebraically.

Thanks for all your help!
-Meghan

Hi Meghan,

I think you have probably done what was expected.

nP2 = n!/(n-2)! = n(n-1)(n-2)!/(n-2)! = n(n-1) = 90

and what remains is to find two consecutive integer with their product being 90.

For the second problem 6Pn = 6 5 4 ... and you continus until you get a product of 720. So for example if the problem had been to find k is 6Pk = 360 then, since 6 5 4 3 = 360 = 6!/2! = 6P4, k = 4.

Penny

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