Question

If $\alpha { = {}^m}{C_2}$, 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:
We will expand the value of $\alpha { = ^m}{C_2}$ using $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Similarly, simplify $^\alpha {C_2}$ and then substitute the value of $\alpha $ in the expression of $^\alpha {C_2}$. Then, use the property of factorial and simplify the expression to get the required answer.

Complete step by step solution:
We are given that, $\alpha { = ^m}{C_2}$
We will expand the expression by using the identity $^n{C_r}$ as $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Hence, $\alpha = \dfrac{{m!}}{{2!\left( {m - 2} \right)!}}$
It is also known that \[n! = n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1\]
Then,
$
  \alpha = \dfrac{{m\left( {m - 1} \right)\left( {m - 2} \right)!}}{{2.1\left( {m - 2} \right)!}} \\
   \Rightarrow \alpha = \dfrac{{m\left( {m - 1} \right)}}{2}
$
We have to find the value of $^\alpha {C_2}$ which is also equal to $\dfrac{{\alpha !}}{{2!\left( {\alpha - 2} \right)!}}$
Also, we can rewrite it as \[\dfrac{{\alpha \left( {\alpha - 1} \right)\left( {\alpha - 2} \right)!}}{{2!\left( {\alpha - 2} \right)}} = \dfrac{{\alpha \left( {\alpha - 1} \right)}}{2}\]
On substituting the values of $\alpha $, we will get,
\[\dfrac{{\dfrac{{m.\left( {m - 1} \right)}}{2}\left( {\dfrac{{m.\left( {m - 1} \right)}}{2} - 1} \right)}}{2} = \dfrac{{m.\left( {m - 1} \right)\left( {m.\left( {m - 1} \right) - 2} \right)}}{8}\]
Solve the brackets.
$
  \dfrac{{\left( {{m^2} - m} \right)\left( {{m^2} - m - 2} \right)}}{8} \\
   = \dfrac{{\left( {{m^2} - m} \right)\left( {{m^2} - 2m + m - 2} \right)}}{8} \\
   = \dfrac{{m\left( {m - 1} \right)\left( {m\left( {m - 2} \right) + 1\left( {m - 2} \right)} \right)}}{8} \\
   = \dfrac{{\left( {m - 2} \right)\left( {m - 1} \right)m\left( {m + 1} \right)}}{8}
$
We will multiply and divide by $\left( {m - 3} \right)!$
 \[\dfrac{{\left( {m - 2} \right)\left( {m - 1} \right)m\left( {m + 1} \right)\left( {m - 3} \right)!}}{{8\left( {m - 3} \right)!}} = \dfrac{{\left( {m + 1} \right)!}}{{8\left( {m - 3} \right)!}}\]
We can write $m - 3$ as $\left( {\left( {m + 1} \right) - 4} \right)$
Then,
\[\dfrac{{\left( {m + 1} \right)!}}{{8\left( {\left( {\left( {m + 1} \right) - 4} \right)} \right)!}}\]
Next, multiply and divide by 3 to make expression 4! In the denominator.
\[\dfrac{{3\left( {m + 1} \right)!}}{{4!\left( {\left( {\left( {m + 1} \right) - 4} \right)} \right)!}} = {3^{m + 1}}{C_4}\]

Hence, option D is the correct answer.

Note:
Students must know how to open expressions like ${}^n{C_r}$. Also, the value of \[n!\] is \[n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1\]. Combination is used to select $r$ objects from $n$ objects when the arrangement of the objects does not matter.