Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:
(a) $\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}$
(b) $n \cdot n-1 C_{r-1}=(n-r+1)^{n} C_{r-1}$
(c) $\frac{{ }^{n} C_{r}}{{ }^{n-1} C_{r-1}}=\frac{n}{r}$
(iv) ${ }^{n} C_{r}+2 \cdot{ }^{n} C_{r-1}+{ }^{n} C_{r-2}={ }^{n+2} C_{r}$.
(a) $\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}$
$\mathrm{LHS}=\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}$
$=\frac{n !}{r !(n-r) !} \times \frac{(r-1) !(n-r+1) !}{n !}$
$=\frac{(n-r+1)(n-r) !(r-1) !}{r(r-1) !(n-r) !}$
$=\frac{n-r+1}{r}=\mathrm{RHS}$
$\therefore \mathrm{LHS}=\mathrm{RHS}$
(b) $\mathrm{LHS}=n .^{n-1} C_{r-1}$
$=\frac{n(n-1) !}{(r-1) !(n-1-r+1) !}$
$=\frac{n !}{(r-1) !(n-r) !}$
$\mathrm{RHS}=(n-r+1)^{n} C_{r}$
$=(n-r+1) \frac{n !}{(r-1) !(n-r+1) !}$
$=(n-r+1) \frac{n !}{(r-1) !(n-r+1)(n-r) !}$
$=\frac{n !}{(r-1) !(n-r) !}$
$\therefore \mathrm{LHS}=\mathrm{RHS}$
(c) $\frac{n_{C_{r}}}{n-1_{C_{r-1}}}=\frac{n}{r}$
$\mathrm{LHS}=\frac{{ }^{n} C_{r}}{{ }^{n-1} C_{r-1}}$
$=\frac{n !}{r !(n-r) !} \times \frac{(r-1) !(n-1-r+1) !}{(n-1) !}$
$=\frac{n(n-1) !}{r(r-1) !(n-r) !} \times \frac{(r-1) !(n-r) !}{(n-1) !}$
$=\frac{n}{r}=\mathrm{RHS}$
$\therefore \mathrm{LHS}=\mathrm{RHS}$
(d) ${ }^{n} C_{r}+2 .{ }^{n} C_{r-1}+{ }^{n} C_{r-2}={ }^{n+2} C_{r}$
$\mathrm{LHS}={ }^{n} C_{r}+2 .^{n} C_{r-1}+{ }^{n} C_{r-2}$
$=\left({ }^{n} C_{r}+{ }^{n} C_{r-1}\right)+\left({ }^{n} C_{r-1}+{ }^{n} C_{r-2}\right)$
$={ }^{n+1} C_{r}+{ }^{n+1} C_{r-1} \quad\left[\because n_{C_{r}}+n_{C_{r, 1}}=n+1_{C_{C}}\right]$
$={ }^{n+2} C_{r} \quad\left[\because n_{C_{r}}+n_{C_{r-1}}=n+1_{C_{r}}\right]$
= RHS
$\therefore \mathrm{LHS}=\mathrm{RHS}$