# If ${}^n{P_7} = 42\left( {{}^n{P_5}} \right)$, then find $n$

Answer

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Hint- Here, we will be using the general formula for permutations.

Given, ${}^n{P_7} = 42\left( {{}^n{P_5}} \right){\text{ }} \to {\text{(1)}}$

Since, we know that the general formula for permutation is given by

Number of ways of arranging $r$ items out of $n$ items is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$

Solving the given equation (1) using above formula, we get

$

\dfrac{{n!}}{{\left( {n - 7} \right)!}} = 42\left[ {\dfrac{{n!}}{{\left( {n - 5} \right)!}}} \right] \\

\Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)\left( {n - 6} \right)\left( {n - 7} \right)!}}{{\left( {n - 7} \right)!}} = 42\left[ {\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)!}}{{\left( {n - 5} \right)!}}} \right] \\

\Rightarrow n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)\left( {n - 6} \right) = 42n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right) \\

\Rightarrow \left( {n - 5} \right)\left( {n - 6} \right) = 42 \Rightarrow {n^2} - 11n + 30 = 42 \Rightarrow {n^2} - 11n - 12 = 0 \\

\Rightarrow {n^2} + n - 12n - 12 = 0 \Rightarrow n\left( {n + 1} \right) - 12\left( {n + 1} \right) = 0 \Rightarrow \left( {n + 1} \right)\left( {n - 12} \right) = 0 \\

\\

$

Either $n = - 1$ or $n = 12$

Since, the value of $n$ should always be positive so we will neglect $n = - 1$.

Therefore, the possible value of $n$ is 12.

Note- In these types of problems we have to check at the end that the values of $n$ we are getting are non-negative. If any value of $n$ comes out to be negative, then that value is not considered because that value is not feasible.

Given, ${}^n{P_7} = 42\left( {{}^n{P_5}} \right){\text{ }} \to {\text{(1)}}$

Since, we know that the general formula for permutation is given by

Number of ways of arranging $r$ items out of $n$ items is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$

Solving the given equation (1) using above formula, we get

$

\dfrac{{n!}}{{\left( {n - 7} \right)!}} = 42\left[ {\dfrac{{n!}}{{\left( {n - 5} \right)!}}} \right] \\

\Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)\left( {n - 6} \right)\left( {n - 7} \right)!}}{{\left( {n - 7} \right)!}} = 42\left[ {\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)!}}{{\left( {n - 5} \right)!}}} \right] \\

\Rightarrow n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)\left( {n - 6} \right) = 42n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right) \\

\Rightarrow \left( {n - 5} \right)\left( {n - 6} \right) = 42 \Rightarrow {n^2} - 11n + 30 = 42 \Rightarrow {n^2} - 11n - 12 = 0 \\

\Rightarrow {n^2} + n - 12n - 12 = 0 \Rightarrow n\left( {n + 1} \right) - 12\left( {n + 1} \right) = 0 \Rightarrow \left( {n + 1} \right)\left( {n - 12} \right) = 0 \\

\\

$

Either $n = - 1$ or $n = 12$

Since, the value of $n$ should always be positive so we will neglect $n = - 1$.

Therefore, the possible value of $n$ is 12.

Note- In these types of problems we have to check at the end that the values of $n$ we are getting are non-negative. If any value of $n$ comes out to be negative, then that value is not considered because that value is not feasible.

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