\hspace{-16}$\bf{(A)} No. of ordered pairs $\bf{(n,r)}$ which satisfy $\bf{\binom{n}{r}=2013}$\\\\\\ (B) No. of ordered pairs $\bf{(n,r)}$ which satisfy $\bf{\binom{n}{r}=2014}$
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3 Answers
2305(A)We try to check for all solutions other than (2013,1)& (2013,2012)
2013 = 3*11*61
Therefore n≥61.(otherwise it can never have the prime factor 61)
We can check that r =0,(61-0),1,(61-1),2,(61-2) does not satisfy the given condition.
For 58≥r≥3.
\inline \dpi{200} \binom{n}{r}\geq \binom{n}{3}\geq \binom{61}{3}> 2013
Hence no other solution can exist.
(B)Similarly for 2014 solutions other than (2014,1)& (2014,2013) does not exist.
1357First one seems to have only two solution
(2013,1); (2013,2012)
Soumyabrata Mondal is there any particular method to do this typ of sum??Upvote·0· Reply ·2013-02-20 04:34:22
1057a little hint can be drawn from the fact that all of non-proper divisors of 2013 are prime(3,11,61).
