Explanation

(a)
Given, $\ ^{n}C_{r} =^{n}C_{r - 1}\$and $\ ^{n}P_{r} = {\ ^{n}P}_{r + 1}$
$\Rightarrow \ \frac{n!}{(n - r)!r!} = \frac{n!}{\begin{array}{r}
(n - r + 1)!(r - 1)! \\
\
\end{array}}$
and $\frac{n!}{(n - r)!} = \frac{n!}{(n - r - 1)!}$
$\Rightarrow n - r + 1 = r$ and $n - r = 1$
$\Rightarrow n - 2r + 1 = 0$ and $n - r - 1 = 0$
On solving, we get
$n = 3,\ r = 2$