Question

If $n= \,^mC_2$, then the value of $^nC_2$ is given by

• A. $3\left(^{m+1}C_{4}\right)$
• B. $^{m-1}C_{4}$
• C. $^{m+1}C_{4}$
• D. $2\left(^{m+2}C_{4}\right)$

Answer 1

The Correct Option is A

$n=^{m}C_{2}=\frac{m\left(m-1\right)}{2}$ Since $m$ and $\left(m - 1\right)$ are two consecutive natural numbers, therefore their product is an even natural number. So $\frac{m\left(m-1\right)}{2}$ is also a natural number. Now $\frac{m\left(m-1\right)}{2}=\frac{m^{2}-m}{2}$ $\therefore \frac{m\left(m-1\right)}{2}C_{2}=\frac{\left(\frac{m^{2}-m}{2}\right)\left(\frac{m^{2}-m}{2}-1\right)}{2}$ $=\frac{m\left(m-1\right)\left(m^{2}-m-2\right)}{8}$ $=\frac{m\left(m-1\right)\left[m^{2}-2m+m-2\right]}{8}$ $=\frac{m\left(m-1\right)\left[m\left(m^{2}-2\right)+1\left(m-2\right)\right]}{8}$ $=\frac{m\left(m-1\right)\left[m\left(m-2\right)\left(m+1\right)\right]}{8}$ $=\frac{3\times\left(m+1\right)m\left(m-1\right)\left(m-2\right)}{4\times3\times2\times1}=3\left(^{m+1}C_{4}\right)$