Question

If we have an expression as $\alpha ={}^{m}{{C}_{2}}$, then ${}^{\alpha }{{C}_{2}}$ is equal to: -
(a) ${}^{m+1}{{C}_{4}}$
(b) ${}^{m-1}{{C}_{4}}$
(c) $3.{}^{m+2}{{C}_{4}}$
(d) $3.{}^{m+1}{{C}_{4}}$

Answer 1

Find the value of $\alpha$ in terms of m by expanding the relation $\alpha ={}^{m}{{C}_{2}}$, using the formula given as: - ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Now, expand the relation: - ${}^{\alpha }{{C}_{2}}$ by using the same formula and substitute the value of $\alpha$ found in terms of m earlier. Simplify the expression to get the answer.

Complete step-by-step solution
Here, we have been provided with the relation $\alpha ={}^{m}{{C}_{2}}$ and then find the value of ${}^{\alpha }{{C}_{2}}$.
Now, we know that ${}^{n}{{C}_{r}}$ is expanded by the formula: - ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, so applying this formula, we get,

& \Rightarrow \alpha ={}^{m}{{C}_{2}} \\\ & \Rightarrow \alpha =\dfrac{m!}{2!\left( m-2 \right)!} \\\ \end{aligned}$$ Now, m! can be written as $$m!=m\times \left( m-1 \right)\times \left( m-2 \right)!$$, so we get, $$\Rightarrow \alpha =\dfrac{m\left( m-1 \right)\left( m-2 \right)!}{2!\left( m-2 \right)!}$$ Cancelling $$\left( m-2 \right)!$$, we get, $$\Rightarrow \alpha =\dfrac{m\left( m-1 \right)}{2!}$$ $$\Rightarrow \alpha =\dfrac{m\left( m-1 \right)}{2}\left( \because 2!=2\times 1=2 \right)$$ - (1) Now, applying the same formula to expand $${}^{\alpha }{{C}_{2}}$$, we get, $$\begin{aligned} & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\alpha !}{2!\left( \alpha -2 \right)!} \\\ & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\alpha \left( \alpha -1 \right)}{2} \\\ \end{aligned}$$ Substituting the value of $$\alpha $$ from equation (1) in the above relation, we get, $$\begin{aligned} & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\dfrac{m\left( m-1 \right)}{2}\left( \dfrac{m\left( m-1 \right)}{2}-1 \right)}{2} \\\ & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( m\left( m-1 \right)-2 \right)}{8} \\\ & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( {{m}^{2}}-m-2 \right)}{8} \\\ \end{aligned}$$ Now, splitting the middle term of the expression $${{m}^{2}}-m-2$$ in the above relation, we get, $$\begin{aligned} & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( {{m}^{2}}-2m+m-2 \right)}{8} \\\ & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( m\left( m-2 \right)+1\left( m-2 \right) \right)}{8} \\\ & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( m+1 \right)\left( m-2 \right)}{8} \\\ \end{aligned}$$ The numerator in the above relation can be arranged as: - $$\Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\left( m+1 \right)m\left( m-1 \right)\left( m-2 \right)}{8}$$ Now, multiplying both the numerator and denominator of the above expression with (m – 3) (m – 4) (m – 5)…….. 3 . 2 . 1, we get, $$\Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\left( m+1 \right)m\left( m-1 \right)\left( m-2 \right)\left( m-3 \right)......3.2.1}{8\times \left( m-3 \right)\left( m-4 \right)\left( m-5 \right)......3.2.1}$$ Now, we have the numerator as the multiplication of terms starting from (m + 1) and ending at 1 and the denominator as the multiplication of terms starting from (m – 3) and ending at 1, so they can be written as factorial like: - $$\Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\left( m+1 \right)!}{8\left( m-3 \right)!}$$ Now, multiplying the numerator and the denominator with 4!, we get, $$\begin{aligned} & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\left( m+1 \right)!}{8\left( m-3 \right)!}\times \dfrac{4!}{4!} \\\ & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{4!}{8}\times \dfrac{\left( m+1 \right)!}{4!\times \left( m-3 \right)!} \\\ \end{aligned}$$ This can be written as: - $$\begin{aligned} & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{4!}{8}\times {}^{m+1}{{C}_{4}} \\\ & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{4\times 3\times 2\times 1}{8}\times {}^{m+1}{{C}_{4}} \\\ & \Rightarrow {}^{\alpha }{{C}_{2}}=3.{}^{m+1}{{C}_{4}} \\\ \end{aligned}$$ **Hence, option (d) is the correct answer.** **Note:** You must remember the expansion formula of the combination expression $${}^{n}{{C}_{r}}$$ to solve the above question. Note that we must arrange the terms properly so that a particular arrangement can be formed and the expression can be written as the factorial of different terms. You may note that we can easily get the correct option by substituting m = 5 in the expression and the options. But this process can only be applied if the options are present just like in the above question.