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

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.

## closed as off-topic by YuiTo Cheng, Thomas Shelby, Jendrik Stelzner, user1551, Parcly TaxelMay 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 $$n=13$$.

• When you solve the last equation, you also get $n = 8$, but this is obviously invalid since $n - 12 \geq 0$. – 1123581321 May 25 at 3:32
• Ah yeah, I'll edit that in – auscrypt May 25 at 3:53

$$\displaystyle \frac{n!}{(n-12)!}=\frac{n!}{(n-10)!}\times6=\frac{n!}{(n-10)(n-11)\times(n-12)!}\times6 \implies 1=\frac{6}{(n-10)(n-11)}$$

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 $$n-10=3$$.

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

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

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

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

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

So to compute nPr we do $$n(n-1)\ldots(n-r+1)$$. We note the following recurence

$$$$nPr = nP(r-1) \times (n-r+1)$$$$

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.