Question

If $^n{C_8}{ = ^n}{C_2}$, then find the value of $^n{C_2}$.

Answer 1

Hint: In this question we have been given an equation in combinations with one value – ‘n’ missing and we have been asked to find the value of $^n{C_2}$. First, find the value of ‘n’ by using certain properties of combinations. Do not expand using the formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$. Once you have found the value of ‘n’, put it in $^n{C_2}$ and expand using the formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$. This will give you your required answer.

Formula used: $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$

Complete step-by-step answer:
We are given an equation in combinations and we have been asked the value of $^n{C_2}$. At first, we will find the value of ‘n’, then using the value of ‘n’, we will find the value of $^n{C_2}$.
We will use a property of combinations which says that if $^n{C_r}{ = ^n}{C_p}$, then either $r = p$ or $r = n - p$. But in this case, $r \ne p$. So, we will try the other property.
It is given that: $^n{C_8}{ = ^n}{C_2}$
By using $r = n - p$, putting $r = 8,p = 2$.
$ \Rightarrow 8 = n - 2$
On solving we have,
$ \Rightarrow n = 10$
Now, we have to find the value of $^n{C_2}$. Putting n = 10, we have ${ \Rightarrow ^{10}}{C_2}$
By using $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ ,
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10!}}{{\left( {10 - 2} \right)!2!}}$
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10!}}{{8!2!}}$
Opening the factorials,
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10 \times 9 \times 8!}}{{8! \times 2 \times 1}}$
On simplifying we will get,
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10 \times 9}}{2}$
${ \Rightarrow ^{10}}{C_2} = 45$

Hence, the value of $^n{C_2}$ is 45.

Note: 1) While expanding $^n{C_8}{ = ^n}{C_2}$, we cannot use the formula of $^n{C_r}$ as it will make an equation in degree 5 or 6, which will make it impossible to solve. This is the reason why we will use properties to solve the question and hence, students must be aware of the properties of combinations.
2) In the step when we were opening the factorials $\left( {^{10}{C_2} = \dfrac{{10!}}{{8!2!}}} \right)$, we opened \[10!\]. Because it is bigger than the factorials in the denominator and can cancel the other factorials if expanded.