Question

Let r and n be positive integers such that 1 ≤ r ≤ n. Then prove the following:

\[\frac{^{n}{}{C}_r}{^{n - 1}{}{C}_{r - 1}} = \frac{n}{r}\]

Answer 1

\[\frac{n_{C_r}}{n - 1_{C_{r - 1}}} = \frac{n}{r}\]

\[LHS = \frac{{}^n C_r}{{}^{n - 1} C_{r - 1}} \]
\[ = \frac{n!}{r! \left( n - r \right)!} \times \frac{\left( r - 1 \right)! \left( n - 1 - r + 1 \right)!}{\left( n - 1 \right)!} \]
\[ = \frac{n \left( n - 1 \right)!}{r \left( r - 1 \right)! \left( n - r \right)!} \times \frac{\left( r - 1 \right)! \left( n - r \right)!}{\left( n - 1 \right)!} \]
\[ = \frac{n}{r} = RHS\]

∴  LHS = RHS