Question

If ${}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{r}}$ , find the value of $r$ .

Answer 1

Hint:For solving this question, we will use the formulas like $r!=r\times \left( r-1 \right)!$ and ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ to write ${}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{5}}$ . After that, we will compare it with the given equation, to get the value of $r$ easily.

Complete step-by-step answer:
Given:
It is given that, if ${}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{r}}$ and we have to find the value of $r$ .
Now, before we proceed we should know the following formulas:
$\begin{align}
  & r!=r\times \left( r-1 \right)!.............\left( 1 \right) \\
 & {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}...............\left( 2 \right) \\
\end{align}$
Now, we will use the above two formulas to solve this question.
First, we will simplify the term on the left-hand side of the equation ${}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{r}}$ , with the help of the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ Then,
$\begin{align}
  & {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{9!}{\left( 9-5 \right)!}+5\times \dfrac{9!}{\left( 9-4 \right)!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{9!}{4!}+5\times \dfrac{9!}{5!} \\
\end{align}$
Now, we will use the formula from the equation (1) to write $5!=5\times 4!$ in the above equation. Then,
$\begin{align}
  & {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{9!}{4!}+5\times \dfrac{9!}{5!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{9!}{4!}+5\times \dfrac{9!}{5\times 4!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{9!}{4!}+\dfrac{9!}{4!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=2\times \dfrac{9!}{4!} \\
\end{align}$
Now, we will multiply by 5 in the numerator and denominator in $2\times \dfrac{9!}{4!}$ , in the above equation. Then,
$\begin{align}
  & {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=2\times \dfrac{9!}{4!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{2\times 5}{5}\times \dfrac{9!}{4!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{10\times 9!}{5\times 4!} \\
\end{align}$
Now, we will use the formula from the equation (1) to write $10\times 9!=10!$ and $5\times 4!=5!$ in the above equation. Then,
$\begin{align}
  & {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{10\times 9!}{5\times 4!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{10!}{5!} \\
\end{align}$
Now, we will write $5!=\left( 10-5 \right)!$ in the above equation. Then,
$\begin{align}
  & {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{10!}{5!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{10!}{\left( 10-5 \right)!} \\
\end{align}$
Now, we will use the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ , from equation (2) to write, $\dfrac{10!}{\left( 10-5 \right)!}={}^{10}{{P}_{5}}$ in the above equation. Then,
$\begin{align}
  & {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}=\dfrac{10!}{\left( 10-5 \right)!} \\
 & \Rightarrow {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{5}}...............\left( 3 \right) \\
\end{align}$
Now, from the above result, we conclude that, ${}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{5}}$ . And it is given that, ${}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{r}}$ so, we can compare the given equation and equation (3). Then,
$\begin{align}
  & {}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{r}}={}^{10}{{P}_{5}} \\
 & \Rightarrow {}^{10}{{P}_{r}}={}^{10}{{P}_{5}} \\
 & \Rightarrow r=5 \\
\end{align}$
Now, from the above result, we conclude that the value of $r$ will be equal to $5$ .
Thus, if ${}^{9}{{P}_{5}}+5{}^{9}{{P}_{4}}={}^{10}{{P}_{r}}$ , then $r=5$ .

Note: Here, the student should first understand the question and then proceed in the right direction to get the correct answer quickly. Moreover, we could have solved the question directly using the formula ${}^{n}{{P}_{r}}+r{}^{n}{{P}_{r-1}}={}^{n+1}{{P}_{r}}$ with the values, $n=9$ and $r=5$ . And in such questions, whenever we got stuck then just try to break every term into factorial terms and proceed further easily.