# Solve the following

Question:

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

Solution:

${ }^{\alpha} C_{2}=\frac{\alpha}{2} \times \frac{(\alpha-1)}{1} \times{ }^{\alpha} C_{0} \quad\left[\because{ }^{n} C_{r}=\frac{n}{r} \cdot{ }^{n-1} C_{r-1}\right]$

$=\frac{1}{2} \alpha(\alpha-1) \quad\left[\because{ }^{n} C_{0}=1\right]$

$=\frac{1}{2}\left[{ }^{m} C_{2}\left({ }^{m} C_{2}-1\right)\right]$

$=\frac{1}{2}\left[\frac{m !}{2 !(m-2) !}\left(\frac{m !}{2 !(m-2) !}-1\right)\right]$

$=\frac{1}{2}\left[\frac{m(m-1)}{2}\left(\frac{m(m-1)}{2}-1\right)\right]$

$=\frac{1}{2}\left[\frac{m(m-1)}{2}\left(\frac{m(m-1)-2}{2}\right)\right]$

$=\frac{1}{8}[m(m-1)\{m(m-1)-2\}]$

$=\frac{1}{8}\left[m^{2}(m-1)^{2}-2 m(m-1)\right]$

$=\frac{1}{8}\left[m^{2}\left(m^{2}+1-2 m\right)-2 m^{2}+2 m\right]$

$=\frac{1}{8}\left[m^{4}+m^{2}-2 m^{3}-2 m^{2}+2 m\right]$

$=\frac{1}{8}\left[m^{4}-2 m^{3}-m^{2}+2 m\right]$

$=\frac{1}{8}\left[\left(m^{2}-2 m\right)\left(m^{2}-1\right)\right]$

$=\frac{1}{8}[m(m-2)(m-1)(m+1)]$

$=\frac{1}{8}(m+1) m(m-1)(m-2)$