Question

If ${}^{n}C_r={}^{n}C_{r-1}$ and ${}^{n}P_r={}^{n}P_{r+1}$ then the value of $n$is

• A. $3$
• B. $4$
• C. $2$
• D. $5$

Answer 1

The correct option is A

$3$

Explanation for the correct option

Step 1: Solve for the relation between $n$ and $r$

Given that ${}^{n}C_r={}^{n}C_{r-1}...\left[1\right]$ and

${}^{n}P_r={}^{n}P_{r+1}...\left[2\right]$

We know that ${}^{n}C_r={}^{n}C_{n-r}...\left[3\right]$

Comparing equations $\left[1\right]$ and $\left[3\right]$ we get

$\begin{array}{crcl}& {}^{n}C_{r-1}& =& {}^{n}C_{n-r}\\ \Rightarrow & r-1& =& n-r\\ \Rightarrow & n& =& 2r-1...\left[4\right]\end{array}$

Considering equation $\left[2\right]$,

$\begin{array}{crcll}& {}^{n}P_r& =& {}^{n}P_{r+1}& \\ \Rightarrow & \frac{n!}{\left(n-r\right)!}& =& \frac{n!}{\left(n-\left(r+1\right)\right)!}& \left[\because {}^{n}P_r=\frac{n!}{\left(n-r\right)!}\right]\\ \Rightarrow & \frac{1}{\left(n-r\right)!}& =& \frac{1}{\left(n-r-1\right)!}& \\ \Rightarrow & \frac{1}{\left(n-r\right)\left(n-r-1\right)!}& =& \frac{1}{\left(n-r-1\right)!}& \\ \Rightarrow & n-r& =& 1& \\ \Rightarrow & n& =& r+1& ...\left[5\right]\end{array}$

Step 2: Solve for the required value

Comparing equations $\left[4\right]$ and $\left[5\right]$, we get,

$\begin{array}{crcl}& 2r-1& =& r+1\\ \Rightarrow & r& =& 2\end{array}$

Substituting the value of $r$ in equation $\left[5\right]$

$\begin{array}{crcl}\therefore & n& =& r+1\\ \Rightarrow & n& =& 2+1\\ \Rightarrow & n& =& 3\end{array}$

Hence the correct option is option(A) i.e. $3$