You can put this solution on YOUR website! Find n so that nC3=nP2
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n!/[(n-3)!*3!] = n!/(n-2)!
Divide both sides by n! to get:
1/[(n-3)!*6] = 1/(n-2)!
Invert to get:
6(n-3)! = (n-2)*(n-3)!
Divide both sides by (n-3)! to get:
6 = (n-2)
n = 8
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Cheers,
Stan H.