Question

# : If ${}^{12}{{P}_{r}}={}^{11}{{P}_{6}}+6\times {}^{11}{{P}_{5}}$, then the value of ‘r’ is ,A.3B.6C.9D.12

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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.

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.