Given,
\(\Rightarrow\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}=\frac{1}{2}\)
\(\Rightarrow\frac{(r+1)!(n-r-1)!}{r!(n-r)!}=\frac{1}{2}\)
\(\Rightarrow\frac{r+1}{n-r}=\frac{1}{2}\)
⇒ = 3 + 2 ...(i)
Similarly,
\(\frac{^nC_{r+1}}{^nC_{r+2}}=\frac{2}{3}\)
\(\Rightarrow \frac{(\frac{n!}{(r+1)!(n-r-1)!})}{(\frac{n!}{(r+2)!(n-r-1)!})}=\frac{2}{3}\)
\(\Rightarrow \frac{(r+2)(n-r-2)}{(r+1)(n-r-1)}=\frac{2}{3}\)
\(\Rightarrow \frac{r+2}{n-r-1}=\frac{2}{3}\)
\(\Rightarrow 3r+6=2n-2r-2 \)
⇒ 2n = 5r + 8
Solving (i) and (ii), we get
6r + 4 = 5r + 8
⇒ r = 4
And n = 14
