+1206 What is the largest integer \(n\) for which \(\binom{8}{3} + \binom{8}{4} = \binom{9}{n}\)?
+118608 \(\binom{8}{3} + \binom{8}{4} = \binom{9}{n}\\ LHS\\ =\binom{8}{3} + \binom{8}{4} \\ =\frac{8!}{3!\;5!}+\frac{8!}{4!\;4!}\\ =\frac{9! }{3!\;5!*9}+\frac{9!}{4!\;4!*9}\\ =\frac{9! }{3!\;4!*5*9}+\frac{9!}{3!*4\;*4!*9}\\ =\frac{9!}{3!4!*9}\left( \frac{1}{5}+\frac{1}{4} \right)\\ =\frac{9!}{3!4!*9}\left( \frac{9}{20} \right)\\ =\frac{9!}{3!4!}\left( \frac{1}{4*5} \right)\\ =\frac{9!}{4!5!}\\ =\binom{9}{4}\quad or \quad \binom{9}{5}\)
So the largest value of n is 5
+4609 There is actually a trick for this! \( {n \choose k} \)+\( {n \choose k+1} \)=\( {n+1 \choose k+1} \), so in this case, we have \( {8 \choose 3} + {8 \choose 4} = {9 \choose 4} \) or \( {9 \choose 5} \). Thus, \(\boxed{n=5}.\)
