If ${}^nP_{12}={}^nP_{10}×6$, than what is $n$? 
If ${}^nP_{12}={}^nP_{10}×6$, than what is $n$?

I am at year 11. I do understand the concept of $^nP_r,{}^nC_r$. Once I know the $n$ I can calculate. I got stuck on this.
 A: By definition,
$$\frac{n!}{(n-12)!}=6\frac{n!}{(n-10)!}$$
Divide by $n!$ and multiply by $(n-10)!$ to get
$$(n-10)(n-11)=6$$
Since $n>10$ else $^nP_{10}$ is bad, the only solution is $n=13$.
A: $\displaystyle \frac{n!}{(n-12)!}=\frac{n!}{(n-10)!}\times6=\frac{n!}{(n-10)(n-11)\times(n-12)!}\times6 \implies 1=\frac{6}{(n-10)(n-11)}$

Apart from using the factorial definition, we can do some counting.
Suppose that I have $n$ objects and I want to arrange $10$ of them in order. There are $P_{10}^n$ ways to do that. Now I want to arrange $12$ of them in order and I find that there are $6\times P_{10}^n$ ways to do that. So, for each arrangement of $10$ objects, I have $6$ ways to put two more objects after the $10$ arranged objects. It means that there are $6$ ways to arrange $2$ of the remaining $n-10$ objects. So, we have $P_2^{n-10}=6$. It is easy to see that $n-10=3$.
A: $^nP_{12} = 6 \ ^nP_{10}$
$\frac{n!}{(n-12)!} = 6\frac{n!}{(n-10)(n-11)(n-12)!}$
${(n-10)(n-11)} = 6$
$(n-10)(n-11) = 2(3)$
$n = 10+3 $ or $n = 11+2 = 13$
A: So to compute nPr we do $n(n-1)\ldots(n-r+1)$. We note the following recurence
\begin{equation}
nPr = nP(r-1) \times (n-r+1)
\end{equation}
If you apply this recurrence twice, you should find a quadratic equation that $n$ must satisfy. When you find the solutions, pick the one such that both $nP12$ and $nP10$ make sense, and you can then verify that the solution is correct.
