# A problem with factorials

I am trying to find a simple generalisable method which can be used to solve questions of the following type:

1) Find $n$ given $\frac{n!}{(n-3)!}=504$.

2) Find $n$ given $\frac{n!}{(n-5)!}=720$.

It is not hard to guess that for $n=9$ works for (1) and $n=6$ works for (2). I arrived at these by somewhat ad hoc methods. I am looking for a neat way to instruct students to solve this type of question and which will work for larger values. I would appreciate any help.

• Thanks for the help everyone. I can defiantly use a lot of these points to help explain. – Number1729 Nov 18 '16 at 9:10

$\frac{n!}{(n-3)!}=504\implies$

$(n-0)(\color\red{n-1})(n-2)=504\implies$

$\color\red{n-1}\approx\sqrt{504}\implies$

$n\approx\sqrt{504}+1\implies$

$n\approx9$

$\frac{n!}{(n-5)!}=720\implies$

$(n-0)(n-1)(\color\red{n-2})(n-3)(n-4)=720\implies$

$\color\red{n-2}\approx\sqrt{720}\implies$

$n\approx\sqrt{720}+2\implies$

$n\approx6$

• I think you meant *bear in mind – RGS Nov 18 '16 at 9:01
• @RSerrao: Thanks. I'm pretty sure that bare is the correct word in this context. There are several debates about it if you look it up on the web. In any case, my answer is not so good IMO, because it doesn't give a general method for answering this type of questions. I should probably fix "$n\approx\sqrt{504}$" to "$n-1\approx\sqrt{504}$", and in the other case denote "$n-2\approx\sqrt{720}$". – barak manos Nov 18 '16 at 9:04
• @barakmanos: don't worry, your method is general, the $k^{th}$ root bounds both $n$ and $n-k+1$, below and above. It is probably faster than trial & error based on prime factorization of the RHS, and I guess that you can make it even tighter (from asymptotics of the root). – Yves Daoust Nov 18 '16 at 9:11
• @YvesDaoust: Yep, but I've fixed it to something a little more more general IMO. Thanks. – barak manos Nov 18 '16 at 9:14

You simplify the ratio as

$$n(n-1)(n-2)$$

Then from the prime factorization of the RHS, you can find a factorization in three successive integers rather easily.

$$504=2^3\cdot3^2\cdot7$$ immediately gives you

$$7\cdot8\cdot9.$$

Similarly

$$2^4\cdot3^2\cdot5\to2\cdot3\cdot4\cdot5\cdot6.$$

Extra hint:

The inequalities $$n-2<\sqrt{n(n-1)(n-2)}<n$$ gives you a fairly good bracketing of $n$:

$$n-2<7.96<n$$ i.e. $n=8$ or $9$. Just try $6\cdot7\cdot8$ and $7\cdot8\cdot9$ (bingo).

Similarly,

$$n-4<3.73<n\to n=3,4,5,6\text{ or }7\to1\cdot2\cdot3\cdot4\cdot5,\ 2\cdot3\cdot4\cdot5\cdot6\text{ (bingo)}.$$ (The bracketing is a little looser because the numbers are small compared to $4$, but the first values can be rejected.)

$$\frac{n!}{(n-3)!}=504$$ $$\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}=504$$ $$n(n-1)(n-2)=504$$ $${\color{Red} n}{\color{blue} {(n-1)} }{\color{Magenta}{(n-2)} }={\color{Red}9 }*\color{blue}8*{\color{Magenta}{7} }$$ it is very clear that the $n=9$.

you can use same idea for the second

The first one is equivalent to $(n-2)(n-1)n=504$. I suppose you meant $n$ is natural number. If that's the case, then we can bound $(n-2)(n-1)n$. Note that for $n \ge 10$ then $(n-2)n(n-1) \ge 8 \cdot 9 \cdot 10>504$.

If $n \le 8$ then $(n-2)(n-1)n \le 6 \cdot 7 \cdot 8<504$. Thus $n=9$.

(2) is similar to (1). You just need to take a guess and prove that it is the only solution like above.

An expression of the type $$\frac{n!}{(n-k)!}$$ is a nice and compact way of writing the product of $k$ consecutive numbers, starting with $n$:

$$\frac{n!}{(n-k)!} = n(n-1)(n-2)(n-3)\cdots(n-k+1)$$

So your students can either try to guess, solve the polynomial equation (this yields a polynomial of degree $k$) or any other neat trick they can think of, depending on the value of $k$.

An easy way of estimating the answer is, for the first one $n(n-1)(n-2)\approx (n-1)^3$ and the second similarly $\approx (n-2)^5$. These approximations will be a little high, which you can see e.g. for the first $$n(n-1)(n-2)=(n-1)\left((n-1)^2-1\right)=(n-1)^3-(n-1)$$So take the relevant root to get the integer to try.