Question 154622
P(n,3)= 2C(n,2)
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n!/(n-3)! = 2[n!/(n-2)!2!]

Cancel the n! and the 2, both of which are common factors to get:

1/(n-3)! = 1/(n-2)!

Invert to get:

(n-3)! = (n-2)!

(n-3)! = (n-2)(n-3)!

Divide both sides by (n-3)!

n-2 = 1
n = 3
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Cheers,
Stan H.