Answer: \({}^{15}C_8=6435\).

Answer: \({}^{15}C_8=6435\).

Use the formula \({}^nC_r=\frac{n!}{r!(n-r)!}\). Then

\({}^nC_4=\frac{n(n-1)(n-2)(n-3)}{24},\)

and

\({}^nC_2=\frac{n(n-1)}{2}.\)

The equation becomes

\(\frac{n(n-1)(n-2)(n-3)}{24}=13\cdot\frac{n(n-1)}{2}.\)

Since \(n\ge4\), divide by \(n(n-1)\):

\(\frac{(n-2)(n-3)}{24}=\frac{13}{2}.\)

So

\((n-2)(n-3)=156.\)

Hence

\(n^2-5n+6=156,\)

so

\(n^2-5n-150=0.\)

Factorise:

\((n-15)(n+10)=0.\)

Since \(n\) is positive, \(n=15\). Therefore

\({}^nC_8={} ^{15}C_8={} ^{15}C_7=6435.\)