Answer
Verified
481.2k+ views
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.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Who was the Governor general of India at the time of class 11 social science CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference Between Plant Cell and Animal Cell