# : If ${}^{12}{{P}_{r}}={}^{11}{{P}_{6}}+6\times {}^{11}{{P}_{5}}$, then the value of ‘r’ is ,

A.3

B.6

C.9

D.12

Answer

Verified

327k+ views

Hint: Use the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ in the given equation and after simplification do the adjustment of multiplying the numerator and denominator by 6 to get the step $\dfrac{12!}{\left( 12-r \right)!}=\dfrac{12}{6}\times \dfrac{11!}{5!}$ and after that simplify the equation to get the answer.

Complete step-by-step answer:

To solve the above problem we will write the given equation first,

${}^{12}{{P}_{r}}={}^{11}{{P}_{6}}+6\times {}^{11}{{P}_{5}}$ ……………………………………….. (1)

To find the value of ‘r’ we have to simplify the solution first and for that we have to know the formula given below,

Formula:

${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$

If we use the above formula in equation (1) we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{\left( 11-6 \right)!}+6\times \dfrac{11!}{\left( 11-5 \right)!}$

Now by further simplification we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{5!}+6\times \dfrac{11!}{6!}$

As we know that 6! Can be expressed as, $6\times 5\times 4\times 3\times 2\times 1$, therefore we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{5!}+6\times \dfrac{11!}{6\times 5\times 4\times 3\times 2\times 1}$

Now we can cancel the 6 from both the numerator and denominator, therefore we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{5!}+\dfrac{11!}{5\times 4\times 3\times 2\times 1}$

As we know that, $5\times 4\times 3\times 2\times 1$ can be expressed as 5! therefore we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{5!}+\dfrac{11!}{5!}$

If we add the terms of the right hand side we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=2\dfrac{11!}{5!}$

Now make the numerator of the right hand side equal to 12! We will do some adjustment in the above equation,

Therefore we will multiply the numerator and denominator of the right hand side by 6, therefore we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=2\times \dfrac{6}{6}\times \dfrac{11!}{5!}$

Multiplying 6 by 2 in the right hand side of the above equation we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{12}{6}\times \dfrac{11!}{5!}$

As we know that $12\times 11!=12!$ and $6\times 5!=6!$ therefore the above equation will become,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{12!}{6!}$

If we observe the above equation then we can say that if we do some adjustment in the above equation then we can get the equation in the form of ${}^{n}{{P}_{r}}$ directly,

Therefore we will replace ‘6’ by (12-6) as both have same values therefore above equation will become,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{12!}{\left( 12-6 \right)!}$

Now to proceed further in the solution we should the formula given below,

$\dfrac{n!}{\left( n-r \right)!}={}^{n}{{P}_{r}}$

By using above formula we can write the above equation as,

$\therefore {}^{12}{{P}_{r}}={}^{12}{{P}_{6}}$

By simply observing the above equation we can easily write the value of ‘r’ as,

r = 6

Therefore if ${}^{12}{{P}_{r}}={}^{11}{{P}_{6}}+6\times {}^{11}{{P}_{5}}$ , then the value of ‘r’ is 6.

Therefore the correct answer is option (b).

Note: After writing the step $\dfrac{12!}{\left( 12-r \right)!}=\dfrac{12}{6}\times \dfrac{11!}{5!}$ you can directly conclude the answer as 6 because it’s just half of 12 and therefore there’s no need to solve further and it will save your time also.

Complete step-by-step answer:

To solve the above problem we will write the given equation first,

${}^{12}{{P}_{r}}={}^{11}{{P}_{6}}+6\times {}^{11}{{P}_{5}}$ ……………………………………….. (1)

To find the value of ‘r’ we have to simplify the solution first and for that we have to know the formula given below,

Formula:

${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$

If we use the above formula in equation (1) we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{\left( 11-6 \right)!}+6\times \dfrac{11!}{\left( 11-5 \right)!}$

Now by further simplification we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{5!}+6\times \dfrac{11!}{6!}$

As we know that 6! Can be expressed as, $6\times 5\times 4\times 3\times 2\times 1$, therefore we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{5!}+6\times \dfrac{11!}{6\times 5\times 4\times 3\times 2\times 1}$

Now we can cancel the 6 from both the numerator and denominator, therefore we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{5!}+\dfrac{11!}{5\times 4\times 3\times 2\times 1}$

As we know that, $5\times 4\times 3\times 2\times 1$ can be expressed as 5! therefore we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{11!}{5!}+\dfrac{11!}{5!}$

If we add the terms of the right hand side we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=2\dfrac{11!}{5!}$

Now make the numerator of the right hand side equal to 12! We will do some adjustment in the above equation,

Therefore we will multiply the numerator and denominator of the right hand side by 6, therefore we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=2\times \dfrac{6}{6}\times \dfrac{11!}{5!}$

Multiplying 6 by 2 in the right hand side of the above equation we will get,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{12}{6}\times \dfrac{11!}{5!}$

As we know that $12\times 11!=12!$ and $6\times 5!=6!$ therefore the above equation will become,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{12!}{6!}$

If we observe the above equation then we can say that if we do some adjustment in the above equation then we can get the equation in the form of ${}^{n}{{P}_{r}}$ directly,

Therefore we will replace ‘6’ by (12-6) as both have same values therefore above equation will become,

$\therefore \dfrac{12!}{\left( 12-r \right)!}=\dfrac{12!}{\left( 12-6 \right)!}$

Now to proceed further in the solution we should the formula given below,

$\dfrac{n!}{\left( n-r \right)!}={}^{n}{{P}_{r}}$

By using above formula we can write the above equation as,

$\therefore {}^{12}{{P}_{r}}={}^{12}{{P}_{6}}$

By simply observing the above equation we can easily write the value of ‘r’ as,

r = 6

Therefore if ${}^{12}{{P}_{r}}={}^{11}{{P}_{6}}+6\times {}^{11}{{P}_{5}}$ , then the value of ‘r’ is 6.

Therefore the correct answer is option (b).

Note: After writing the step $\dfrac{12!}{\left( 12-r \right)!}=\dfrac{12}{6}\times \dfrac{11!}{5!}$ you can directly conclude the answer as 6 because it’s just half of 12 and therefore there’s no need to solve further and it will save your time also.

Last updated date: 03rd Jun 2023

•

Total views: 327k

•

Views today: 2.83k

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE