Permtations and Combinations value of n

[Louise Johnson]

Question
Find the value of n if:

\(\displaystyle \L\\{}_nC_2 = {}_nP_1\)
some of the professional people on this forum are using C(n,2)=P(n,1) Perhaps it is easier to understand rather than tex

This was was easy enough to find the answer of n=3 by just plugging in a couple of different numbers randomly. Can anyone tell me the proper way to find the value of n in this question?
Thank you
Louise

 

[pka]

That is an odd problem. Because P(n,1)=n and so \(\displaystyle n = \frac{{n!}}{{\left( {n - 2} \right)!\left( {2!} \right)}} = \frac{{n\left( {n - 1} \right)}}{2}\).

Solve for n.