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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.

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closed as off-topic by YuiTo Cheng, Thomas Shelby, Jendrik Stelzner, user1551, Parcly Taxel May 25 at 23:10

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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 $n=13$.

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  • $\begingroup$ When you solve the last equation, you also get $n = 8$, but this is obviously invalid since $n - 12 \geq 0$. $\endgroup$ – 1123581321 May 25 at 3:32
  • $\begingroup$ Ah yeah, I'll edit that in $\endgroup$ – auscrypt May 25 at 3:53
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$\displaystyle \frac{n!}{(n-12)!}=\frac{n!}{(n-10)!}\times6=\frac{n!}{(n-10)(n-11)\times(n-12)!}\times6 \implies 1=\frac{6}{(n-10)(n-11)}$


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 $n-10=3$.

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$^nP_{12} = 6 \ ^nP_{10}$

$\frac{n!}{(n-12)!} = 6\frac{n!}{(n-10)(n-11)(n-12)!}$

${(n-10)(n-11)} = 6$

$(n-10)(n-11) = 2(3)$

$n = 10+3 $ or $n = 11+2 = 13$

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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.

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