Question

if $n={}^m{C}_2$ then ${}^n{C}_2$ equal to

• A. ${}^{m+1}C_4$
• B. ${}^{m-1}C_4$
• C. ${}^{m+2}C_4$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for correct option:

Binomial coefficient simplification:

Given, $n={}^{m}C_2$

$$\begin{gathered} \therefore {}^{m}C_2=\frac{m!}{(2!(m-2)!)} \\ =\frac{m(m-1)}{2} \end{gathered}$$

Now ,

$$\begin{gathered} {}^{n}C_2=\frac{n!}{2!(n-2)!} \\ =\frac{n(n-1)}{2} \\ =\frac{\left(\frac{m(m-1)}{2}\right)\left(\frac{m(m-1)}{2}-1\right)}{2} \\ =\left(\frac{m(m-1)}{4}\right)\left(\frac{m(m-1)}{2}-1\right) \\ =\left(\frac{m(m-1)}{4}\right)\left(\frac{m^2-m-2}{2}\right) \\ =\frac{m(m-1)(m-2)(m+1)}{2\times 4} \\ =3\times \frac{m(m-1)(m-2)(m+1)\left((m-3)!\right)}{1·2·3·4·\left((m-3)!\right)} \\ =3\times {}^{m+1}{C}_4\left[\because {}^n{C}_r=\frac{n!}{\left(r!\right)\left(n-r\right)!}\right] \end{gathered}$$n,

Hence option D is correct.