Question

The least positive integer $n$ such that $(n-1)C_5+(n-1)C_6=n{C}_7$ then the value $n-7$ is

Answer 1

Simplifying ${}^{n-1}{C}_5+{}^{n-1}{C}_6<{}^n{C}_7$

$\frac{\left(n-1\right)!}{5!\left(n-6\right)!}+\frac{(n-1)!}{6!\left(n-7\right)!}=\frac{n!}{7!\left(n-7\right)!}$

$\frac{6\left(n-1\right)!+\left(n-6\right)\left(n-1\right)!}{6!\left(n-6\right)\left(n-7\right)!}=\frac{n!}{7!\left(n-7\right)!}$

$\frac{\left(n-1\right)!\left(6+n-6\right)}{6!\left(n-6\right)\left(n-7\right)!}=\frac{n!}{7!\left(n-7\right)!}$

$\frac{1}{n-6}=\frac{1}{7}$

$n-6=7$

$n=13$

Therefore, the value of $n-7$ is,

$n-7=13-7$

$=6$