Question

If we are given the ratio of permutation as ${}^{n}{{P}_{4}}:{}^{n}{{P}_{5}}=1:2$, then $n$ is:
1. 4
2. 5
3. 6
4. 7

Answer 1

Hint: For solving this question you should know about the general formula of permutations. Here in this question we will set the given terms according to the formula of permutation and then we will compare that with the given ratio and find the value of $n$. The formula of permutations is given by $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.

Complete step-by-step solution:
According to the question given to us we are asked to find the value of $n$ if ${}^{n}{{P}_{4}}:{}^{n}{{P}_{5}}=1:2$. So, if we take the expression given to us, then,
${}^{n}{{P}_{4}}:{}^{n}{{P}_{5}}=1:2\ldots \ldots \ldots \left( i \right)$
Since we know that the general formula for permutations is given by, that is the number of ways of arranging $r$ items out of $n$ items is denoted as $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.
Solving the given equation (i) using the above formula, we will get as follows,
$\begin{align}
  & \dfrac{\dfrac{n!}{\left( n-4 \right)!}}{\dfrac{n!}{\left( n-5 \right)!}}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{\left( n-5 \right)!}{\left( n-4 \right)!}=\dfrac{1}{2} \\
\end{align}$
Simplifying further we will get as follows,
$\begin{align}
  & \dfrac{n!}{\left( n-4 \right)\left( n-5 \right)!}:\dfrac{n!}{\left( n-5 \right)!}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{n!}{\left( n-4 \right)\left( n-5 \right)!}\times \dfrac{\left( n-5 \right)!}{n!}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{1}{\left( n-4 \right)}=\dfrac{1}{2} \\
 & \Rightarrow n-4=2 \\
 & \Rightarrow n=6 \\
\end{align}$
Hence, we get the value of $n$ as 6 and so the correct answer is option 3.

Note: While solving such types of questions, we have to always check at the end if the final value that we are getting for $n$ is a non-negative value or not. If by any chance you get the value of $n$ as a negative one, then that value is not considered because that value is not a feasible value. So, always remember to check for the same in such questions.