Question

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

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

Answer 1

\[\frac{{}^n C_r}{{}^n C_{r - 1}} = \frac{n - r + 1}{r}\]

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

∴\[LHS = RHS\]