1

Possible Duplicate:
How to reverse the $n$ choose $k$ formula?

Recently saw an ad for a salad bar in Spain offering 1000 different combinations. Now, this could be a salad containing three parts, with a choice of ten ingredients for each part, but this was probably just an advertisement effort. There is one more (unlikely) option that I found interesting:

The salad bar could potentially have $n$ ingredients, from which you could choose only $m$, if the following condition held:

$${m \choose n } = \frac{n!}{m!(n-m)!} = 1000$$

And more generally, for a given $N$, do there exist $m,n$ such that: $${m \choose n } = N$$ Is there a general method of finding such a pair? is such a pair unique?

Community
  • 1
Nathaniel Bubis
  • 33,018
  • Your binomial coefficient is upside-down. Also, note the trivial solutions $n=1000, m\in\lbrace 1,999\rbrace$. – Sean Eberhard Oct 01 '12 at 21:40
  • 4
    On the other hand, Erdos proved that there are no solutions to ${n\choose m}=x^3$ for $1<m\leq n/2$. –  Oct 01 '12 at 21:43
  • 1
    The linked question doesn't address the uniqueness problem, which as far as I know is a hard open problem; of course if ${m \choose n} = N$ then ${N \choose 1} = {N \choose N-1} = {m \choose m-n} = N$, but classifying all of the other nontrivial repetitions among binomial coefficients is open, I think. – Qiaochu Yuan Oct 01 '12 at 21:44
  • @Byron That's a lovely sledge-hammer-upon-pea solution. More pedestrianly, certainly $\binom{n}{m}\neq 1000$ for $n>1000$ and $m\neq 0,1000$, so at least there is a brute-force solution. – Sean Eberhard Oct 01 '12 at 21:47
  • 1
    Apart from obvious patterns of non-uniqueness, there is for example $\binom{16}{2}=\binom{10}{3}=120$. – André Nicolas Oct 01 '12 at 21:50
  • I don't imagine there is any legal obstacle to their advertising 1000 combinations if the correct number is really $14\choose 4$, or even if it were $20\choose 10$. – MJD Oct 01 '12 at 22:02
  • @MJD - this isn't about the legality, it's about the math inspired by it :) – Nathaniel Bubis Oct 04 '12 at 23:20
  • In related news, I recently visited a gelato store that advertises 218 flavors. I counted the tubs and found that there were actually 224 different flavors on sale. I asked why all the signage proclaimed 218; the attendant said “in case we run out of some”. – MJD Jan 12 '15 at 18:16

0 Answers0