If \({}^{n}C_{9}={}^{n}C_{8}\), find \({}^{n}C_{17}\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
Choosing \(r\) things out of \(n\) is the same as leaving out the other \(n-r\), so \({}^{n}C_{r}={}^{n}C_{n-r}\). Hence, if \({}^{n}C_{x}={}^{n}C_{y}\) with \(x\ne y\), then \(x+y=n\).
The two lower numbers are different, so they add up to \(n\): \(n=9+8=17\)
\({}^{17}C_{17}=1\): there is only one way to choose all 17