Ordered pair (n,r) in Binomial Coefficients

[man111]

Number of positive integer ordered pairs [tex](n,r)[/tex] for which [tex]\displaystyle \binom{n}{r} = 2016[/tex]

[Knowledge Greedy]

Отговор: [tex](2016; \,\ 1)[/tex] , [tex](2016; \,\ 2015)[/tex] , [tex](64; \,\ 2)[/tex] , [tex](64; \,\ 62)[/tex]
Но доказателството, че това са всички, ми убягва.

[man111]

To Knowledge Greedy would you like to explain me how can i get it

Thanks

[nevrodermit]

[tex]2016=2^5.3^2.7[/tex]
Clearly, if [tex](n,r)[/tex] is a solution then [tex](n,n-r)[/tex] is a solution too from the properties of the binomial coefficient. So we can consider only the case when [tex]r\le n-r \Leftrightarrow 2r \le n[/tex].

Case 1: [tex]r=1[/tex]. Then [tex]n=2016[/tex] so [tex](2016,1)[/tex] is a solution, and hence [tex](2016, 2015)[/tex] is.
Case 2: [tex]r=2[/tex]. Then [tex]n(n-1)=2.2016[/tex]. Solving this quadratic equation gives us [tex]n=64, r=2[/tex] and so [tex](64,2), (64,62)[/tex] are solutions.
Case 3: [tex]r=3[/tex]. Then [tex]n(n-1)(n-2)=6.2016=2^6.3^3.7[/tex]. Since exactly one among [tex]n, n-1, n-2[/tex] can be divisible by [tex]3[/tex] we get that exactly one of them is divisible by [tex]27[/tex] and hence [tex]n\ge 27k, k\in Z\Rightarrow 6.2016=n(n-1)(n-2)\ge 27^3[/tex] which is not true.
Case 4: [tex]r>3[/tex]. Note that [tex]{n \choose r } \le {n \choose r+1} \Leftrightarrow \frac{n!}{r!(n-r)!}\le \frac{n!}{(r+1)!(n-r-1)!} \Leftrightarrow r+1\le n-r\Leftrightarrow 2r\le n-1[/tex]. Now we have to consider two cases:
Case 4.1: [tex]2r=n[/tex]. Then [tex](2r)!=2016(r!)^2[/tex]. From the Bertrand's postulate for [tex]r>3[/tex] there is a prime [tex]p[/tex] such that [tex]r<p<2r[/tex] and so [tex]p|2016, 3<p\Rightarrow p=7 \Rightarrow 3<r<7\Rightarrow r\in \{4,5,6\}[/tex] with corresponding values for [tex]n\in \{8,10,12\}[/tex]. One can check that none of these pairs is a solution with simple calculation.
Case 4.2: [tex]2r\le n-1[/tex]. Then using the established inequality we get [tex]2016 = {n \choose r}\ge {n \choose 3}\ge\frac{27^3}{6}[/tex] which is not true and hence no solutions in this case.

[man111]

Thanks nevrodermit for nice solution