Question

Find the value of n such that,
(i) ${}^{n}{{P}_{5}}=42\times {}^{n}{{P}_{3}},n>4$
(ii) $\dfrac{{}^{n}{{P}_{4}}}{{}^{n-1}{{P}_{4}}}=\dfrac{5}{3},n>4$

Answer 1

Hint: We will first start by using the property that ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ in part (i) and solve it to find the value of n. Then we will similarly solve the part (ii) by using the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ and simplify.

Complete step-by-step answer:
Now, we will first solve the part (i) ${}^{n}{{P}_{5}}=42\times {}^{n}{{P}_{3}},n>4$
Now, we know that the identity ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.
So, using this we have,
$\dfrac{n!}{\left( n-5 \right)!}=42\times \dfrac{n!}{\left( n-3 \right)!}$
Now, we will cross multiply the expression and cancel n! in numerator and denominator,
$\begin{align}
  & \dfrac{n!\times \left( n-3 \right)!}{\left( n-5 \right)!\times n!}=42 \\
 & \dfrac{\left( n-3 \right)!}{\left( n-5 \right)!}=42 \\
\end{align}$
Now, we know that,
$\begin{align}
  & \left( n-3 \right)!=\left( n-5 \right)!\left( n-4 \right)\left( n-3 \right) \\
 & =\dfrac{\left( n-5 \right)!\left( n-4 \right)\left( n-3 \right)}{\left( n-5 \right)!}=42 \\
 & =\left( n-4 \right)\left( n-3 \right)=42 \\
\end{align}$
Now, we will expand the left side of the equation. So, we have,
$\begin{align}
  & {{n}^{2}}-7n+12=42 \\
 & {{n}^{2}}-7n+12-42=0 \\
 & {{n}^{2}}-7n-30=0 \\
\end{align}$
Now, we will split the middle term as a sum of 30.
$\begin{align}
  & {{n}^{2}}-10n+3n-30=0 \\
 & n\left( n-10 \right)+3\left( n-10 \right)=0 \\
 & \left( n+3 \right)\left( n-10 \right)=0 \\
 & either\ n+3=0\ or\ n-10=0 \\
 & \Rightarrow n=-3\ or\ n=10 \\
\end{align}$
Since, n > 4 has been given to us, therefore $n\ne -3\ and\ n=10$.
Now, in part (ii) we have $\dfrac{{}^{n}{{P}_{4}}}{{}^{n-1}{{P}_{4}}}=\dfrac{5}{3},n>4$.
Similarly, we will use ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. So, we have,
$\begin{align}
  & \dfrac{\dfrac{n!}{\left( n-4 \right)!}}{\dfrac{\left( n-1 \right)!}{\left( n-5 \right)!}}=\dfrac{5}{3} \\
 & \dfrac{n!\times \left( n-5 \right)!}{\left( n-4 \right)!\times \left( n-1 \right)!}=\dfrac{5}{3} \\
\end{align}$
Now, we will write,
\[\begin{align}
  & n!=\left( n-1 \right)!\times n \\
 & \left( n-4 \right)!=\left( n-5 \right)!\times \left( n-4 \right) \\
 & \dfrac{\left( n-1 \right)!\times n\times \left( n-5 \right)!}{\left( n-5 \right)!\times \left( n-4 \right)\times \left( n-1 \right)!}=\dfrac{5}{3} \\
 & \dfrac{n}{n-4}=\dfrac{5}{3} \\
\end{align}\]
Now, on cross – multiply we have,
$\begin{align}
  & 3n=5\left( n-4 \right) \\
 & 3{{n}}-5n-20=0 \\
 & \Rightarrow 3n=5n-20 \\
 & 20=5n-3n \\
 & 2n=20 \\
 & n=10 \\
\end{align}$
So, the value of n is 10 for part (ii).

Note: It is important to note that while solving the first part we have used the factorization method to solve the quadratic equation and in the second part we have a linear equation to find the value of n. Also, it is important to note that we have used the fact $n!=\left( n-1 \right)n$ to solve both parts.