How to calculate n if we have r and nPr? (permutations)

Question

There is this practice problem "How many students are in a class if we can pick 3 delegates in 6545 ways?" I've tried solving this by using the permutations formula but I got left with something that looks like a cubic equation and we didn't even study those yet. So I probably got something wrong.

Answer 1

You have ${n \choose 3} = \frac{n(n-1)(n-2)}{6} = 6545$, so $n(n-1)(n-2) = 39270$. The solution is an integer, so just try a few values of $n$. You quickly find that $n = 35$ works.

You could also factorize $39270 = 5 \cdot 6 \cdot 7 \cdot 11 \cdot 17$ and try some divisors. Clearly it can't be any of the prime divisors, so you try multiples. $5 \cdot 6 = 30$ doesn't work, so you try $5 \cdot 7 = 35$, which works. I doubt factorizing will save you time, but it is a little nicer than brute force.

Answer 2

If you are allowed to use a calculator, then calculate $\sqrt[3]{6545 \cdot 6} \approx 34$

$6545 \cdot6 = 39270$ so therefore we know that one of $n, n-1, n-2$ must be a multiple of $5$. Indeed, when $n=35$, $n \cdot (n-1) \cdot (n-2) = 39270$. Walla! :)

Answer 3

Since you know that $n$ has to be an integer, you can use the rational root theorem.

We have ${n\choose3}=6545,$ so that $$n(n-1)(n-2)-39720=0.$$ The constant term is $-39720=-2\cdot3\cdot5\cdot7\cdot11\cdot17$ and by the rational root theorem, an integer solution must divide this. You can check all the factors in turn, of course, but notice that we have $39720=n(n-1)(n-2)\approx n^3,$ so that $n^3$ is a bit more than $40,000.$ Since $30^3=27,000$ and $40^3=64,000,$ looking at the prime factors of $39720$ suggests that $35=5\cdot7$ is a good candidate to try first.

EDIT

Okay, I got lucky, because $5$ and $7$ are next to one another in the factor list, but the only other possible candidates are $34=2\cdot17$ and $33=3\cdot11,$ so it wasn't all that lucky.

Answer 4

If $${n \choose k} = x$$ then for large $n$ and small $k$ you can say ${n \choose k}$ is slightly less than $\frac{1}{k!}\left(n - \frac{k-1}{2}\right)^k$ and so $$n = \left\lceil \frac{k-1}{2}+\sqrt[k]{k! x}\right\rceil$$

In your example with $x=6545$ and $k=3$ this gives $n=\left\lceil 1+\sqrt[3]{6\times 6545}\right\rceil = \left\lceil 34.99\right\rceil = 35$ and checking ${35 \choose 3}$ confirms it