# ordered pairs in binomial coefficients

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?

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Factoring $2013$ was a good idea. For (2), you'll want to start by factoring $2014$. – Gerry Myerson Jan 22 '13 at 6:30

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

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 $r=1$

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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