solve P(k,3) = 2C(k,2) for k, where k > 0

[Clifford]

Solve P(k,3) = 2C(k,2) for k, where k > 0.

k! / (k-3)! = 2k! / 2!(k-2)!

k! / (k-3)! = k! / (k-2)!

k(k-1)(k-2) = k(k-1)

(k-2) = 0

k = 2

When substituting k = 2 back into the left side and right side at the top it doesn't work out. What am I doing wrong?

 

[pka]

There is silly mistake.
It should be (k-2)=1 not 0.

 

[Clifford]

Right, silly me cancelling out doesn't = 0. Thanks, now k = 3 and it fits the equations fine.