If (^{n}C_{r}=^{n}C_{r+1}), what is the relation between (n) and (r)?

Correct answer: \(n=2r+1\)

Given \(^{n}C_r=^{n}C_{r+1}\). Using the ratio of consecutive combinations, \(\frac{^{n}C_{r+1}}{^{n}C_r}=\frac{n-r}{r+1}\). For the two combinations to be equal, this ratio must be 1. Hence \(n-r=r+1\), so \(n=2r+1\). The relation \(n=2r\) does not generally make two consecutive combinations equal. Exam tip: use the symmetry \(^{n}C_r=^{n}C_{n-r}\) to write \(r+1=n-r\) directly.