# How do I solve for n in this permutation question?

I have the following question:

Solve for n:

$$_nP_3 = 6_{n-1}P_2$$

I don't know how I should begin to tackle this problem? Any tips/help would be appreciated.

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Note that we can "expand" with how $_nP_m$ is defined: $$\dfrac{n!}{(n-m)!}= \dfrac{n(n-1)(n - 2)\cdots}{(n-m)(n-m-1)\cdots} = n(n-1) \cdots (n-m + 1)$$

\begin{align} _nP_3 &= n(n-1)(n-2) \\ \\ 6(_{n-1}P_2) & = 6(n - 1)(n-2)\end{align}

Now, since both expressions are given as equal to one another, we must have $n = 6$.

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You must convert both sides of the equation into the equivalent of a permutation.

So, nP3 would become n!/(n-3)!

the other side would be 6((n-1)!/(n-3)!)

now you just rearrange the equations and solve for n:

the (n-3)! cancel on both sides so you are left with n! = 6(n-1)!

you notice that 6(n-1)! is = n! and thus n = 6

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\begin{eqnarray*} _{n}{\rm P}_{3} & = & 6\left(_{n-1}{\rm P}_{2}\right)\\ \frac{n!}{\left(n-3\right)!} & = & 6\frac{\left(n-1\right)!}{\left(\left(n-1\right)-2\right)!}\\ \frac{n\times\left(n-1\right)!}{\left(n-3\right)!} & = & 6\frac{\left(n-1\right)!}{\left(n-3\right)!}\\ n & = & 6 \end{eqnarray*}

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