query: Given a specific rational number, is there a way to find an n and k for the binomial coefficient that will evaluate to it?
positive_passages.docid: 143132
positive_passages.text: How to reverse the $n$ choose $k$ formula?
negative_passages.docid: 56317
negative_passages.text: Intuitive explanation of binomial coefficient formula
negative_passages.docid: 25673
negative_passages.text: Number after binomial coefficient
negative_passages.docid: 142303
negative_passages.text: Alternating sum of binomial coefficients multiplied by (1/k+1)
negative_passages.docid: 98082
negative_passages.text: Another combinatorics problem: $\sum\limits_{k = 0}^n (-1)^k \binom{2n-k}k2^{2n-2k}=2n+1$
negative_passages.docid: 151384
negative_passages.text: Summation of series involving binomial coefficients and polynomial of degree at most n-1
negative_passages.docid: 39957
negative_passages.text: Product of binomial coefficient as a basis
negative_passages.docid: 122024
negative_passages.text: Asymptotic of a sum involving binomial coefficients
negative_passages.docid: 131559
negative_passages.text: Lower Bound of Central Binomial Coefficients
negative_passages.docid: 150438
negative_passages.text: Binomial theorem for non integers ? O_o ??
negative_passages.docid: 157552
negative_passages.text: Find the coefficient of $x^n$ simply by using binomial theorem
negative_passages.docid: 126587
negative_passages.text: Divisiblity of binomial coefficients
negative_passages.docid: 10191
negative_passages.text: Binomial theorem: divisibility by $n^2$
negative_passages.docid: 43188
negative_passages.text: The $n^{th}$ root of the geometric mean of binomial coefficients.
negative_passages.docid: 63893
negative_passages.text: Prime dividing the binomial coefficients
negative_passages.docid: 91648
negative_passages.text: Binomial coefficient proof
negative_passages.docid: 104704
negative_passages.text: Evaluate $\sum\limits_{k=0}^n \binom{n}{k}$ combinatorially