Question

Prove that:
(i) ${}^{n}{{P}_{n}}={}^{n}{{P}_{n-1}}$
(ii) ${}^{n}{{P}_{r}}=n\cdot {}^{n-1}{{P}_{r-1}}$
(iii) ${}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}={}^{n}{{P}_{r}}$

Answer 1

Hint: For solving this question, we will use the formulas like ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ , $r!=r\times \left( r-1 \right)!$ and $0!=1$ to simplify the term on the left-hand side in each of the given part. Then, we solve further and prove the desired result easily.

Complete step-by-step answer:
Given:
We have to prove the following equations:
(i) ${}^{n}{{P}_{n}}={}^{n}{{P}_{n-1}}$
(ii) ${}^{n}{{P}_{r}}=n\cdot {}^{n-1}{{P}_{r-1}}$
(iii) ${}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}={}^{n}{{P}_{r}}$
Now, before we proceed we should know the following formulas and results:
$\begin{align}
  & {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}............\left( 1 \right) \\
 & r!=r\times \left( r-1 \right)!..........\left( 2 \right) \\
 & 0!=1........................\left( 3 \right) \\
\end{align}$
Now, we will solve for each part separately using the above formulas.
(i) ${}^{n}{{P}_{n}}={}^{n}{{P}_{n-1}}$
Now, we will use the formula from the equation (1) to simplify the term ${}^{n}{{P}_{n}}$ in the above equation. Then,
$\begin{align}
  & {}^{n}{{P}_{n}}=\dfrac{n!}{\left( n-n \right)!} \\
 & \Rightarrow {}^{n}{{P}_{n}}=\dfrac{n!}{0!} \\
\end{align}$
Now, we will put $0!=1$ from equation (3) in the above equation. Then,
${}^{n}{{P}_{n}}=\dfrac{n!}{1}$
Now, we will write $1=1!$ in the above equation. Then,
$\begin{align}
  & {}^{n}{{P}_{n}}=\dfrac{n!}{1} \\
 & \Rightarrow {}^{n}{{P}_{n}}=\dfrac{n!}{1!} \\
\end{align}$
Now, we will write $1!=\left( n-\left( n-1 \right) \right)!$ in the above equation. Then,
$\begin{align}
  & {}^{n}{{P}_{n}}=\dfrac{n!}{1!} \\
 & \Rightarrow {}^{n}{{P}_{n}}=\dfrac{n!}{\left( n-\left( n-1 \right) \right)!} \\
\end{align}$
Now, we will use the formula from the equation (1) to write $\dfrac{n!}{\left( n-\left( n-1 \right) \right)!}={}^{n}{{P}_{n-1}}$ in the above equation. Then,
$\begin{align}
  & {}^{n}{{P}_{n}}=\dfrac{n!}{\left( n-\left( n-1 \right) \right)!} \\
 & \Rightarrow {}^{n}{{P}_{n}}={}^{n}{{P}_{n-1}} \\
\end{align}$
Hence, proved.

(ii) ${}^{n}{{P}_{r}}=n\cdot {}^{n-1}{{P}_{r-1}}$
Now, we will use the formula from the equation (1) to simplify the term ${}^{n}{{P}_{r}}$ in the above equation. Then,
\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
Now, we will use the formula from the equation (2) to write $n!=n\times \left( n-1 \right)!$ in the above equation. Then,
\[\begin{align}
  & {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} \\
 & \Rightarrow {}^{n}{{P}_{r}}=\dfrac{n\times \left( n-1 \right)!}{\left( n-r \right)!} \\
\end{align}\]
Now, we will write $\left( n-r \right)=\left( n-1-\left( r-1 \right) \right)$ in the above equation. Then,
\[\begin{align}
  & {}^{n}{{P}_{r}}=\dfrac{n\times \left( n-1 \right)!}{\left( n-r \right)!} \\
 & \Rightarrow {}^{n}{{P}_{r}}=\dfrac{n\times \left( n-1 \right)!}{\left( n-1-\left( r-1 \right) \right)!} \\
 & \Rightarrow {}^{n}{{P}_{r}}=n\times \dfrac{\left( n-1 \right)!}{\left( n-1-\left( r-1 \right) \right)!} \\
\end{align}\]
Now, we will use the formula from the equation (1) to write \[\dfrac{\left( n-1 \right)!}{\left( n-1-\left( r-1 \right) \right)!}={}^{n-1}{{P}_{r-1}}\] in the above equation. Then,
\[\begin{align}
  & {}^{n}{{P}_{r}}=n\times \dfrac{\left( n-1 \right)!}{\left( n-1-\left( r-1 \right) \right)!} \\
 & \Rightarrow {}^{n}{{P}_{r}}=n\times {}^{n-1}{{P}_{r-1}} \\
 & \Rightarrow {}^{n}{{P}_{r}}=n\cdot {}^{n-1}{{P}_{r-1}} \\
\end{align}\]
Hence, proved.

(iii) ${}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}={}^{n}{{P}_{r}}$
Now, we will use the formula from the equation (1) to simplify the term ${}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}$ in the above equation. Then,$\begin{align}
  & {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left( n-1 \right)!}{\left( n-1-r \right)!}+r\times \dfrac{\left( n-1 \right)!}{\left( n-1-\left( r-1 \right) \right)!} \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left
( n-1 \right)!}{\left( n-r-1 \right)!}+r\times \dfrac{\left( n-1 \right)!}{\left( n-r \right)!} \\
\end{align}$
Now, we will use the formula from the equation (2) to write $\left( n-r \right)!=\left( n-r \right)\times \left( n-r-1 \right)!$ in the above equation. Then,
$\begin{align}
  & {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!}+r\times \dfrac{\left( n-1 \right)!}{\left( n-r \right)!} \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!}+r\times \dfrac{\left( n-1 \right)!}{\left( n-r \right)\times \left( n-r-1 \right)!} \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!}+\dfrac{r}{\left( n-r \right)}\times \dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!} \\
\end{align}$
Now, we will take $\dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!}$ common from each term in the above equation. Then,
$\begin{align}
  & {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!}+\dfrac{r}{\left( n-r \right)}\times \dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!} \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!}\times \left[ 1+\dfrac{r}{\left( n-r \right)} \right] \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!}\times \left[ \dfrac{n-r+r}{n-r} \right] \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{\left( n-1 \right)!}{\left( n-r-1 \right)!}\times \dfrac{n}{\left( n-r \right)} \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{n\times \left( n-1 \right)!}{\left( n-r \right)\times \left( n-r-1 \right)!} \\
\end{align}$
Now, we will use the formula from the equation (2) to write $n\times \left( n-1 \right)!=n!$ and $\left( n-r \right)\times \left( n-r-1 \right)!=\left( n-r \right)!$ in the above equation. Then,
$\begin{align}
  & {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{n\times \left( n-1 \right)!}{\left( n-r \right)\times \left( n-r-1 \right)!} \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{n!}{\left( n-r \right)!} \\
\end{align}$
Now, we will use the formula from the equation (1) to write $\dfrac{n!}{\left( n-r \right)!}={}^{n}{{P}_{r}}$ in the above equation. Then,
$\begin{align}
  & {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}=\dfrac{n!}{\left( n-r \right)!} \\
 & \Rightarrow {}^{n-1}{{P}_{r}}+r\cdot {}^{n-1}{{P}_{r-1}}={}^{n}{{P}_{r}} \\
\end{align}$
Hence, proved.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to prove the desired result quickly in a correct way. After that, we should apply formulas like ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ and $r!=r\times \left( r-1 \right)!$ correctly in a proper sequence and solve without any mistake to prove the desired result easily. Moreover, we could have also proved the desired result by simplifying the term on the left-hand side and right-hand side separately using formulas like ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ and $r!=r\times \left( r-1 \right)!$ .