Question 176974
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.