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# Find the value of $^r{C_5}$ if $^{18}{C_r}{ = ^{18}}{C_{r + 2}}$.

Last updated date: 16th Jul 2024
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Hint: The formula for $^n{C_r}$ is $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Use this formula to find out the value of $r$. And then put the value of $r$ in $^r{C_5}$.

According to the question, it is given that $^{18}{C_r}{ = ^{18}}{C_{r + 2}}$.
We know that, $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Using this formula we’ll get:
${ \Rightarrow ^{18}}{C_r}{ = ^{18}}{C_{r + 2}} \\ \Rightarrow \dfrac{{18!}}{{r!\left( {18 - r} \right)!}} = \dfrac{{18!}}{{\left( {r + 2} \right)!\left( {18 - r - 2} \right)!}} \\ \Rightarrow \dfrac{1}{{r!\left( {18 - r} \right)!}} = \dfrac{1}{{\left( {r + 2} \right)!\left( {16 - r} \right)!}} \\ \Rightarrow r!\left( {18 - r} \right)! = \left( {r + 2} \right)!\left( {16 - r} \right)! \\$

We know that, $n! = n\left( {n - 1} \right)\left( {n - 2} \right).....3 \times 2 \times 1$. Using this, we’ll get:
$\Rightarrow r!\left( {18 - r} \right)\left( {18 - r - 1} \right)\left( {18 - r - 2} \right)! = \left( {r + 2} \right)\left( {r + 1} \right)r!\left( {16 - r} \right)! \\ \Rightarrow \left( {18 - r} \right)\left( {17 - r} \right)\left( {16 - r} \right)! = \left( {r + 2} \right)\left( {r + 1} \right)\left( {16 - r} \right)! \\ \Rightarrow \left( {18 - r} \right)\left( {17 - r} \right) = \left( {r + 2} \right)\left( {r + 1} \right) \\ \Rightarrow 306 - 18r - 17r + {r^2} = {r^2} + r + 2r + 2 \\ \Rightarrow 306 - 35r = 3r + 2 \\ \Rightarrow 38r = 304 \\ \Rightarrow r = 8 \\$
So, the value of $r$ is 8.

We have to find out the value of $^r{C_5}$. Putting $r = 8$, we’ll get:
${ \Rightarrow ^r}{C_5}{ = ^8}{C_5}$
Using formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we’ll get:
${ \Rightarrow ^8}{C_5} = \dfrac{{8!}}{{5! \times 3!}} \\ { \Rightarrow ^8}{C_5} = \dfrac{{8 \times 7 \times 6 \times 5!}}{{5! \times 6}} \\ { \Rightarrow ^8}{C_5} = 56 \\$
Thus, the value of $^r{C_5}$ is 56.

Note: This question can be solved by another method as:
We know that if $^n{C_a}{ = ^n}{C_b}$ then either $a = b$ or $a + b = n$ must be true.
So for $^{18}{C_r}{ = ^{18}}{C_{r + 2}}$, we have:
$\Rightarrow r = r + 2{\text{ or }}r + r + 2 = 18$
First condition is not true. So we have:
$\Rightarrow r + r + 2 = 18 \\ \Rightarrow 2r + 2 = 18 \\ \Rightarrow 2r = 16 \\ \Rightarrow r = 8 \\$
We have calculated the value of $r$. While putting it in $^r{C_5}$ we will get the same result.