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Question: If \(\alpha { = {}^m}{C_2}\), then \({}^\alpha {C_2}\) is equal to A. \(^{m + 1}{C_4}\) B. \(^{m...

If α=mC2\alpha { = {}^m}{C_2}, then αC2{}^\alpha {C_2} is equal to
A. m+1C4^{m + 1}{C_4}
B. m−1C4^{m - 1}{C_4}
C. 3m+2C4{3^{m + 2}}{C_4}
D. 3m+1C4{3^{m + 1}}{C_4}

Explanation

Solution

We will expand the value of α=mC2\alpha { = ^m}{C_2} using nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}. Similarly, simplify αC2^\alpha {C_2} and then substitute the value of α\alpha in the expression of αC2^\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, α=mC2\alpha { = ^m}{C_2}
We will expand the expression by using the identity nCr^n{C_r} as n!r!(n−r)!\dfrac{{n!}}{{r!\left( {n - r} \right)!}}
Hence, α=m!2!(m−2)!\alpha = \dfrac{{m!}}{{2!\left( {m - 2} \right)!}}
It is also known that n!=n.(n−1).(n−2).....3.2.1n! = n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1
Then,
α=m(m−1)(m−2)!2.1(m−2)! ⇒α=m(m−1)2 \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 αC2^\alpha {C_2} which is also equal to α!2!(α−2)!\dfrac{{\alpha !}}{{2!\left( {\alpha - 2} \right)!}}
Also, we can rewrite it as α(α−1)(α−2)!2!(α−2)=α(α−1)2\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 α\alpha , we will get,
m.(m−1)2(m.(m−1)2−1)2=m.(m−1)(m.(m−1)−2)8\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.
(m2−m)(m2−m−2)8 =(m2−m)(m2−2m+m−2)8 =m(m−1)(m(m−2)+1(m−2))8 =(m−2)(m−1)m(m+1)8 \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 (m−3)!\left( {m - 3} \right)!
(m−2)(m−1)m(m+1)(m−3)!8(m−3)!=(m+1)!8(m−3)!\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−3m - 3 as ((m+1)−4)\left( {\left( {m + 1} \right) - 4} \right)
Then,
(m+1)!8(((m+1)−4))!\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.
3(m+1)!4!(((m+1)−4))!=3m+1C4\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 nCr{}^n{C_r}. Also, the value of n!n! is n.(n−1).(n−2).....3.2.1n.\left( {n - 1} \right).\left( {n - 2} \right).....3.2.1. Combination is used to select rr objects from nn objects when the arrangement of the objects does not matter.