help

n = 5

8C3 + 8C4 =9C5

56   +  70   = 126 

$\binom{8}{3} + \binom{8}{4} = \binom{9}{n}\\ LHS\\ =\binom{8}{3} + \binom{8}{4} \\ =\frac{8!}{3!\;5!}+\frac{8!}{4!\;4!}\\ =\frac{9! }{3!\;5!*9}+\frac{9!}{4!\;4!*9}\\ =\frac{9! }{3!\;4!*5*9}+\frac{9!}{3!*4\;*4!*9}\\ =\frac{9!}{3!4!*9}\left( \frac{1}{5}+\frac{1}{4} \right)\\ =\frac{9!}{3!4!*9}\left( \frac{9}{20} \right)\\ =\frac{9!}{3!4!}\left( \frac{1}{4*5} \right)\\ =\frac{9!}{4!5!}\\ =\binom{9}{4}\quad or \quad \binom{9}{5}$

So the largest value of n is 5

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, $\boxed{n=5}.$