Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  1. Find the number of ordered pairs $(n,r)$ which satisfy $\binom{n}{r} = 2013$.
  2. Find the number of ordered pairs $(n,r)$ which satisfy $\binom{n}{r} = 2014$.

My Attempt for $(1)$:

By simple guessing, we can find two solutions:

$$ \binom{n}{r} = \binom{2013}{1}=\binom{2013}{2013-1}=\binom{2013}{2012} $$

So two solutions are $(2013,1),(2013,2012)$.

We also know that $\binom{n}{r} = 3\times 11 \times 61$.

How can I calculate the remaining ordered pairs $(n,r)$ from this point?

share|cite|improve this question
Factoring $2013$ was a good idea. For (2), you'll want to start by factoring $2014$. – Gerry Myerson Jan 22 '13 at 6:30
up vote 3 down vote accepted

$$\binom n {r+1}\ge \binom nr \iff \frac{\binom n {r+1}}{\binom nr}\ge 1\iff\frac{n-r}{r+1}\ge1\iff r\le \frac{n-1}2$$

So, $$\binom n r\le \binom n {r+1}\iff r\le \frac{n-1}2$$ and $$\binom n r\ge\binom n {r+1}\iff r\ge\frac{n-1}2$$

For any integer $u,\binom n1=u\implies n=u$ will always have a solution in integers.

For $r=2,\binom n2=\frac{n(n-1)}{2}=2013\iff n^2-n-2\cdot2013=0$ but the discriminant $1+4\cdot2\cdot2013=16105$ is not a perfect square, hence we don't have any rational solution here.

For $r=3,\binom n3=\frac{n(n-1)(n-2)}{1\cdot2\cdot3},$

one of the term in the numerator $n-s$ (say,) where $0\le s\le 2$ is divisible by $61$

So, $n-s=61m,n=61m+s$ for some integer $m$ then $n-t\text{( where $0\le s\le 2$)}\ge 61m-2\ge 59m$ for $m\ge 1$

So, $\binom n3\ge \frac{(59m)^3}{1\cdot2\cdot3}>2013$ for $m\ge1$

Now, $\binom n{r+1}\ge \binom n3$ for $\frac{n-1}2\ge r\ge 3\implies \binom nr>2013$for $\frac{n-1}2\ge r\ge 3$

also $\binom n{n-3}\le \binom nr$ for $\frac{n-1}2\le r\le n-3\implies \binom nr>2013$ for $\frac{n-1}2\le r\le n-3$

$\implies \binom nr>2013$ for $3\le r\le n-3$

As $\binom nr=\binom n{n-r},$ the only other solution is $r=n-1$ corresponding to $r=1$

share|cite|improve this answer
Thanks lab bhattacharjee – juantheron Jan 22 '13 at 17:10
@juantheron, my pleasure. But, I think there should be some smarter way. – lab bhattacharjee Jan 22 '13 at 17:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.