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Question
If α = mC2, then find the value of αC2.
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Solution
\[{}^\alpha C_2 = \frac{\alpha}{2} \times \frac{(\alpha - 1)}{1} \times^\alpha C_0\] [∵\[{}^n C_r = \frac{n}{r} . {}^{n - 1} C_{r - 1}\]]
\[ = \frac{1}{2} \left[ \frac{m!}{2! \left( m - 2 \right)!} \left( \frac{m!}{2! \left( m - 2 \right)!} - 1 \right) \right]\]
\[ = \frac{1}{2} \left[ \frac{m \left( m - 1 \right)}{2} \left( \frac{m \left( m - 1 \right)}{2} - 1 \right) \right]\]
\[ = \frac{1}{2} \left[ \frac{m\left( m - 1 \right)}{2} \left( \frac{m \left( m - 1 \right) - 2}{2} \right) \right]\]
\[ = \frac{1}{8} \left[ m \left( m - 1 \right) \left\{ m \left( m - 1 \right) - 2 \right\} \right]\]
\[ = \frac{1}{8} \left[ m^2 \left( m - 1 \right)^2 - 2m \left( m - 1 \right) \right]\]
\[ = \frac{1}{8} \left[ m^2 \left( m^2 + 1 - 2m \right) - 2 m^2 + 2m \right]\]
\[ = \frac{1}{8} \left[ m^4 + m^2 - 2 m^3 - 2 m^2 + 2m \right]\]
\[ = \frac{1}{8} \left[ m^4 - 2 m^3 - m^2 + 2m \right]\]
\[ = \frac{1}{8} \left[ \left( m^2 - 2m \right) \left( m^2 - 1 \right) \right]\]
\[ = \frac{1}{8} \left[ m \left( m - 2 \right) \left( m - 1 \right) \left( m + 1 \right) \right]\]
\[ = \frac{1}{8} \left( m + 1 \right) m \left( m - 1 \right) \left( m - 2 \right)\]
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