Question

If ${}^{10}{C_r} = {}^{10}{C_{r + 2}}$ then ${}^5{C_r}$ equals
$\left( a \right){\text{ 120}}$
$\left( b \right){\text{ 10}}$
$\left( c \right){\text{ 360}}$
$\left( d \right){\text{ 5}}$

Answer 1

Hint: So for solving this question we will use the combination formula which is given by ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and from it we will find the value for the $r$ . For this, we will first substitute the value of $r$ in ${}^5{C_r}$ and then find the value for it. And in this way, we will solve this problem.

Formula used:
The combination means selecting the items from the sample and is called combination, the formula of combination is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Here, $n$, will be the total number of items in the sample
$r$, will be the number of items to be selected from the sample.

Complete step by step solution:
So in the question, it is given that the ${}^{10}{C_r} = {}^{10}{C_{r + 2}}$
Therefore by using the formula of combination, we will expand the above combination on that formula, we get
$ \Rightarrow \dfrac{{10!}}{{r!\left( {10 - r} \right)!}} = \dfrac{{10!}}{{\left( {r + 2} \right)!\left( {10 - (r + 2)} \right)!}}$
Since the numerator is the same on both the side so it will cancel out each other the equation we get is\[ \Rightarrow \dfrac{1}{{r!\left( {10 - r} \right)!}} = \dfrac{1}{{\left( {r + 2} \right)!\left( {10 - r - 2} \right)!}}\]
Now on doing the cross-multiplication, we get
$ \Rightarrow \left( {r + 2} \right)!\left( {8 - r} \right)! = r!\left( {10 - r} \right)!$
On expanding it, we get
$ \Rightarrow \left( {r + 2} \right)\left( {r + 1} \right)r!\left( {8 - r} \right)! = r!\left( {10 - r} \right)\left( {9 - r} \right)\left( {8 - r} \right)!$
Again canceling the same term, we get
$ \Rightarrow \left( {r + 2} \right)\left( {r + 1} \right) = \left( {10 - r} \right)\left( {9 - r} \right)$
Now on multiplying it, we get
$ \Rightarrow {r^2} + 3r + 2 = 90 - 10r - 9r + {r^2}$
And on solving it, we get
$ \Rightarrow 3r + 2 = 90 - 19r$
And from this, we will get
$ \Rightarrow 22r = 88$
On solving for the value of $r$ , we get
$ \Rightarrow r = 4$ , and we will name it equation $1$
So, now we will calculate ${}^5{C_r}$ , on substituting the value of $r$ and using the combination formula we get
$ \Rightarrow \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}}$
And on solving it, we get
$ \Rightarrow \dfrac{{5!}}{{4!1!}}$
On expanding the above combination we get
$ \Rightarrow \dfrac{{5 \times 4!}}{{4!}}$
So the same term will cancel each other and, we get
$ \Rightarrow 5$
Therefore, ${}^{10}{C_r} = {}^{10}{C_{r + 2}}$ then ${}^5{C_r}$ equals $5$.

Hence, the option $\left( d \right)$ is correct.

Note:
One thing we should always keep in mind, that after expanding the formula of combination, we should try to minimize the calculation to the least and cancel the common terms. Since proceeding without cancellations may increase the complexity of the solution.