Question
If $\alpha={^\text{m}}\text{C}_{2},$ then find the value of ${^{\alpha}}\text{C}_{2}.$
✓
Answer
We have,
$\alpha={^\text{m}}\text{C}_{2}=\frac{\text{m}(\text{m}-1)}{2}$
${^{\alpha}}\text{C}_{2}=\frac{\alpha(\alpha-1)}{2}$
$=\frac{\Big(\frac{\text{m}(\text{m}-1)}{2}\Big)\Big(\frac{\text{m}(\text{m}-1)}{2}-1\Big)}{2}$
$=\frac{\text{m}(\text{m}-1)(\text{m}^{2}-\text{m}-2)}{2\times2\times2}$
$=\frac{\text{m}(\text{m}-1)(\text{m}+1)(\text{m}-2)}{8}$
$=\frac{\text{m}(\text{m}-1)(\text{m}+1)(\text{m}-2)}{4\times2}$
Multiplying with 3, numerator and denominator to make 4.
$=\frac{\text{m}(\text{m}+1)\text{m}(\text{m}-1)(\text{m}-2)}{4.3.2.1}$
$=\frac{3(\text{m}+1)\text{m}(\text{m}-1)(\text{m}-2)}{4!}$
$=3.{^{\text{m+1}}}\text{C}_{4}$
$\alpha={^\text{m}}\text{C}_{2}=\frac{\text{m}(\text{m}-1)}{2}$
${^{\alpha}}\text{C}_{2}=\frac{\alpha(\alpha-1)}{2}$
$=\frac{\Big(\frac{\text{m}(\text{m}-1)}{2}\Big)\Big(\frac{\text{m}(\text{m}-1)}{2}-1\Big)}{2}$
$=\frac{\text{m}(\text{m}-1)(\text{m}^{2}-\text{m}-2)}{2\times2\times2}$
$=\frac{\text{m}(\text{m}-1)(\text{m}+1)(\text{m}-2)}{8}$
$=\frac{\text{m}(\text{m}-1)(\text{m}+1)(\text{m}-2)}{4\times2}$
Multiplying with 3, numerator and denominator to make 4.
$=\frac{\text{m}(\text{m}+1)\text{m}(\text{m}-1)(\text{m}-2)}{4.3.2.1}$
$=\frac{3(\text{m}+1)\text{m}(\text{m}-1)(\text{m}-2)}{4!}$
$=3.{^{\text{m+1}}}\text{C}_{4}$
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