Question

If \[n{ = ^m}{C_2}\] then the value of \[^n{C_2}\] is given by
A. \[^{m + 1}{C_4}\]
B. \[^{m - 1}{C_4}\]
C. \[^{m + 2}{C_4}\]
D. \[{3.^{m + 1}}{C_4}\]

Answer 1

Hint: We use the formula of combinations and expand the given term according to the formula. Write the value of n in terms of m. Similarly open the value of the required combination using the formula and substitute the value of n obtained from the previous combination formula.
* Formula of combination is given by\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\]

Step-By-Step answer:
We are given that \[n{ = ^m}{C_2}\]
Use the combination formula to open the value of n
\[ \Rightarrow n = \dfrac{{m!}}{{(m - 2)!2!}}\]
Now we can expand the numerator using factorial formula
\[ \Rightarrow n = \dfrac{{m(m - 1)(m - 2)!}}{{(m - 2)!2!}}\]
Cancel same factors from numerator and denominator of the fraction
\[ \Rightarrow n = \dfrac{{m(m - 1)}}{{2!}}\]
Also we know \[2! = 2\]
\[ \Rightarrow n = \dfrac{{m(m - 1)}}{2}\] … (1)
Now we have to calculate the value of \[^n{C_2}\]
Use the combination formula to open the value
\[{ \Rightarrow ^n}{C_2} = \dfrac{{n!}}{{(n - 2)!2!}}\]
Now we can expand the numerator using factorial formula
\[{ \Rightarrow ^n}{C_2} = \dfrac{{n(n - 1)(n - 2)!}}{{(n - 2)!2!}}\]
Cancel same factors from numerator and denominator of the fraction
\[{ \Rightarrow ^n}{C_2} = \dfrac{{n(n - 1)}}{{2!}}\]
Also we know \[2! = 2\]
\[{ \Rightarrow ^n}{C_2} = \dfrac{{n(n - 1)}}{2}\] … (2)
Substitute the value of ‘n’ from equation (1) in equation (2)
\[{ \Rightarrow ^n}{C_2} = \dfrac{{\left\{ {\dfrac{{m(m - 1)}}{2}} \right\}\left\{ {\dfrac{{m(m - 1)}}{2} - 1} \right\}}}{2}\]
Take LCM in the numerator of the fraction
\[{ \Rightarrow ^n}{C_2} = \dfrac{{\left\{ {\dfrac{{m(m - 1)}}{2}} \right\}\left\{ {\dfrac{{m(m - 1) - 2}}{2}} \right\}}}{2}\]
Write the fraction in simpler form
\[{ \Rightarrow ^n}{C_2} = \dfrac{{\left\{ {(m)(m - 1)} \right\}\left\{ {m(m - 1) - 2)} \right\}}}{{2 \times 2 \times 2}}\]
Calculate the second bracket
\[{ \Rightarrow ^n}{C_2} = \dfrac{{\left\{ {(m)(m - 1)} \right\}\left\{ {{m^2} - m - 2)} \right\}}}{{2 \times 2 \times 2}}\]
Factorize the quadratic equation in second bracket
\[{ \Rightarrow ^n}{C_2} = \dfrac{{\left\{ {(m)(m - 1)} \right\}\left\{ {{m^2} - 2m + m - 2)} \right\}}}{{2 \times 2 \times 2}}\]
Collect the factors in second bracket
\[{ \Rightarrow ^n}{C_2} = \dfrac{{\left\{ {(m)(m - 1)} \right\}\left\{ {m(m - 2) + 1(m - 2)} \right\}}}{{2 \times 2 \times 2}}\]
\[{ \Rightarrow ^n}{C_2} = \dfrac{{\left\{ {(m)(m - 1)} \right\}\left\{ {(m - 2)(m + 1)} \right\}}}{{2 \times 2 \times 2}}\]
Write the numerator of the fraction in descending order of terms
\[{ \Rightarrow ^n}{C_2} = \dfrac{{(m + 1)(m)(m - 1)(m - 2)}}{{4 \times 2}}\]
Multiply both numerator and denominator by 3
\[{ \Rightarrow ^n}{C_2} = \dfrac{{3(m + 1)(m)(m - 1)(m - 2)}}{{4 \times 3 \times 2 \times 1}}\]
Now we can write the denominator using factorial formula
\[{ \Rightarrow ^n}{C_2} = \dfrac{{3(m + 1)(m)(m - 1)(m - 2)}}{{4!}}\] … (3)
Also, since there are 4 terms in the numerator, we can write \[^{m + 1}{C_4}\] as we know first multiple is \[m + 1\]
We know \[^{m + 1}{C_4} = \dfrac{{(m + 1)!}}{{(m + 1 - 4)!4!}}\] and on expansion we get \[{ \Rightarrow ^{m + 1}}{C_4} = \dfrac{{(m + 1)(m)(m - 1)(m - 2)(m - 3)!}}{{(m - 3)!4!}}\]. Cancel same factors and we get \[{ \Rightarrow ^{m + 1}}{C_4} = \dfrac{{3(m + 1)(m)(m - 1)(m - 2)}}{{4!}}\]
Substitute value of \[^{m + 1}{C_4}\] in RHS of equation (3)
\[{ \Rightarrow ^n}{C_2} = {3^{m + 1}}{C_4}\]

\[\therefore \]Option D is correct.

Note: Many students get confused when the terms in the numerator start from \[m + 1\] instead of m, keep in mind the term from which the expansion starts is the superscript in the combination representation. Also open the factorial one by one to avoid confusion in calculation.