Question

Find value of n if ${}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right)$.

Answer 1

Hint: We know that permutation of ${}^{n}{{P}_{r}}$ can be given, by using the formula, ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. So, by comparing it with ${}^{n}{{P}_{4}}$ and ${}^{n}{{P}_{3}}$, then equating them with each other and using the formula, $\left( n-r \right)!=\left( n-r \right)\left( n-\left( r+1 \right) \right)!$ wherever needed, we will find the value of n.

Complete step-by-step solution -
In question we are given that ${}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right)$ and we are asked to find the value of n, so first of all we will compare it with general expression i.e. ${}^{n}{{P}_{r}}$ and we know that its value in form of permutation can be given by,
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ ……………(i)
Now, on comparing ${}^{n}{{P}_{4}}$ with expression (i) and we can say that for ${}^{n}{{P}_{4}}$, $r=4$ and $n=n$, so now we can write ${}^{n}{{P}_{4}}$ as,
${}^{n}{{P}_{4}}=\dfrac{n!}{\left( n-4 \right)!}$ ………….(ii)
In the same way, on comparing ${}^{n}{{P}_{3}}$ with ${}^{n}{{P}_{r}}$, we can say that for ${}^{n}{{P}_{3}}$, $r=3$ and $n=n$, so now, we can write ${}^{n}{{P}_{3}}$ as,
${}^{n}{{P}_{4}}=\dfrac{n!}{\left( n-3 \right)!}$ ……………….(iii)
Now, in question it is given that ${}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right)$, so on substituting the value of ${}^{n}{{P}_{4}}$ and ${}^{n}{{P}_{3}}$ from expression (ii) and (iii) we will get,
$\dfrac{n!}{\left( n-4 \right)!}=5\left( \dfrac{n!}{\left( n-3 \right)!} \right)$
On further simplifying we will get,
$\Rightarrow \dfrac{n!}{\left( n-4 \right)!}=\dfrac{5n!}{\left( n-3 \right)!}$
Now, the expansion of $\left( n-3 \right)!$, can be given as, $\left( n-3 \right)!=\left( n-3 \right)\left( n-4 \right)!$, by using the formula $\left( n-r \right)!=\left( n-r \right)\left( n-\left( r+1 \right) \right)!$, so on again substituting this value in expression and canceling $n!$ in numerator as they are same, we will get,
$\Rightarrow \dfrac{1}{\left( n-4 \right)!}=\dfrac{5}{\left( n-3 \right)\left( n-4 \right)!}$
Now, on cancelling similar terms i.e. $\left( n-4 \right)!$, we will get,
$\Rightarrow 1=\dfrac{5}{n-3}$
$\Rightarrow 1\left( n-3 \right)=5$
$\Rightarrow n-3=5$
$\Rightarrow n=5+3=8$
Thus, we can say that value of n in ${}^{n}{{P}_{4}}=5\left( {}^{n}{{P}_{3}} \right)$ is 8.

Note: Instead of applying the formula, $\left( n-r \right)!=\left( n-r \right)\left( n-\left( r+1 \right) \right)!$, in $\left( n-3 \right)!$, students might apply in $\left( n-4 \right)!$, such as $\left( n-4 \right)!=\left( n-4 \right)\left( n-\left( 4+1 \right) \right)!$, which will be wrong and students won't be able to simplify the equation. So, students must use formulas properly and avoid the mistakes to get answers.