Answer: \(n=14\).

Answer: \(n=14\).

Start with

\(65\,{}^nC_5=2(n-1)\,{}^{n+1}C_6.\)

Using factorial notation,

\({}^nC_5=\frac{n!}{(n-5)!5!}\)

and

\({}^{n+1}C_6=\frac{(n+1)!}{(n-5)!6!}.\)

Substitute these into the equation:

\(65\cdot\frac{n!}{(n-5)!5!} =2(n-1)\cdot\frac{(n+1)!}{(n-5)!6!}.\)

Divide the right-hand side by the left-hand combinatorial factor:

\(65=\frac{2(n-1)(n+1)!5!}{n!6!}.\)

Since \(\frac{(n+1)!}{n!}=n+1\) and \(\frac{5!}{6!}=\frac16\),

\(65=\frac{2(n-1)(n+1)}{6}.\)

So

\(65=\frac{n^2-1}{3}.\)

Hence

\(n^2-1=195,\)

so

\(n^2=196.\)

Since \(n\) is positive,

\(n=14.\)