# Let r and n be positive integers such that 1 ≤ r ≤ n.

Question:

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) $\frac{{ }^{n} C_{r}}{{ }^{n-1} C_{r-1}}=\frac{n}{r}$

(d) ${ }^{n} C_{r}+2 \cdot{ }^{n} C_{r-1}+{ }^{n} C_{r-2}={ }^{n+2} C_{r}$.

Solution:

(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

$\therefore \mathrm{LHS}=\mathrm{RHS}$