Question

If α = ^mC₂, then find the value of ^αC₂.

Answer 1

\[{}^\alpha C_2 = \frac{\alpha}{2} \times \frac{(\alpha - 1)}{1} \times^\alpha C_0\]  [∵\[{}^n C_r = \frac{n}{r} . {}^{n - 1} C_{r - 1}\]]

\[{}^n C_r = \frac{n}{r} . {}^{n - 1} C_{r - 1}\]  [∵\[{}^n C_0 = 1\]]

\[= \frac{1}{2} \left[ {}^m C_2 \left( {}^m C_2 - 1 \right) \right]\]
\[ = \frac{1}{2} \left[ \frac{m!}{2! \left( m - 2 \right)!} \left( \frac{m!}{2! \left( m - 2 \right)!} - 1 \right) \right]\]
\[ = \frac{1}{2} \left[ \frac{m \left( m - 1 \right)}{2} \left( \frac{m \left( m - 1 \right)}{2} - 1 \right) \right]\]
\[ = \frac{1}{2} \left[ \frac{m\left( m - 1 \right)}{2} \left( \frac{m \left( m - 1 \right) - 2}{2} \right) \right]\]
\[ = \frac{1}{8} \left[ m \left( m - 1 \right) \left\{ m \left( m - 1 \right) - 2 \right\} \right]\]
\[ = \frac{1}{8} \left[ m^2 \left( m - 1 \right)^2 - 2m \left( m - 1 \right) \right]\]
\[ = \frac{1}{8} \left[ m^2 \left( m^2 + 1 - 2m \right) - 2 m^2 + 2m \right]\]
\[ = \frac{1}{8} \left[ m^4 + m^2 - 2 m^3 - 2 m^2 + 2m \right]\]
\[ = \frac{1}{8} \left[ m^4 - 2 m^3 - m^2 + 2m \right]\]
\[ = \frac{1}{8} \left[ \left( m^2 - 2m \right) \left( m^2 - 1 \right) \right]\]
\[ = \frac{1}{8} \left[ m \left( m - 2 \right) \left( m - 1 \right) \left( m + 1 \right) \right]\]
\[ = \frac{1}{8} \left( m + 1 \right) m \left( m - 1 \right) \left( m - 2 \right)\]