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

Hint: 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,
\[\begin{align}
  & \Rightarrow \alpha ={}^{m}{{C}_{2}} \\
 & \Rightarrow \alpha =\dfrac{m!}{2!\left( m-2 \right)!} \\
\end{align}\]
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,
\[\begin{align}
  & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\alpha !}{2!\left( \alpha -2 \right)!} \\
 & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\alpha \left( \alpha -1 \right)}{2} \\
\end{align}\]
Substituting the value of \[\alpha \] from equation (1) in the above relation, we get,
\[\begin{align}
  & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\dfrac{m\left( m-1 \right)}{2}\left( \dfrac{m\left( m-1 \right)}{2}-1 \right)}{2} \\
 & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( m\left( m-1 \right)-2 \right)}{8} \\
 & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( {{m}^{2}}-m-2 \right)}{8} \\
\end{align}\]
Now, splitting the middle term of the expression \[{{m}^{2}}-m-2\] in the above relation, we get,
\[\begin{align}
  & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( {{m}^{2}}-2m+m-2 \right)}{8} \\
 & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( m\left( m-2 \right)+1\left( m-2 \right) \right)}{8} \\
 & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{m\left( m-1 \right)\left( m+1 \right)\left( m-2 \right)}{8} \\
\end{align}\]
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,
\[\begin{align}
  & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{\left( m+1 \right)!}{8\left( m-3 \right)!}\times \dfrac{4!}{4!} \\
 & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{4!}{8}\times \dfrac{\left( m+1 \right)!}{4!\times \left( m-3 \right)!} \\
\end{align}\]
This can be written as: -
\[\begin{align}
  & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{4!}{8}\times {}^{m+1}{{C}_{4}} \\
 & \Rightarrow {}^{\alpha }{{C}_{2}}=\dfrac{4\times 3\times 2\times 1}{8}\times {}^{m+1}{{C}_{4}} \\
 & \Rightarrow {}^{\alpha }{{C}_{2}}=3.{}^{m+1}{{C}_{4}} \\
\end{align}\]
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.