A problem with factorials

Question

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.

Answer 1

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

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

$\color\red{n-1}\approx\sqrt[3]{504}\implies$

$n\approx\sqrt[3]{504}+1\implies$

$n\approx9$

---

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

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

$\color\red{n-2}\approx\sqrt[5]{720}\implies$

$n\approx\sqrt[5]{720}+2\implies$

$n\approx6$

Answer 2

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

you can use same idea for the second

Answer 3

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.)

Answer 4

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.

Answer 5

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 $.

Answer 6

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.