Answer

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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 $^r{C_5}$.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

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