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

$3$

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$4$

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$2$

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$5$

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Solution

The correct option is A
$3$

Explanation for the correct option
Step 1: Solve for the relation between $n$ and $r$
Given that ${}^{n}C_{r} = {}^{n}C_{r - 1} . . . [1]$ and
${}^{n}P_{r} = {}^{n}P_{r + 1} . . . [2]$
We know that ${}^{n}C_{r} = {}^{n}C_{n - r} . . . [3]$
Comparing equations $[1]$ and $[3]$ we get
$\begin{array}{crcl} {}^{n}C_{r - 1} & = & {}^{n}C_{n - r} \\ \Rightarrow & r - 1 & = & n - r \\ \Rightarrow & n & = & 2 r - 1 . . . [4] \end{array}$
Considering equation $[2]$ ,
$\begin{array}{crcll} {}^{n}P_{r} & = & {}^{n}P_{r + 1} \\ \Rightarrow & \frac{n!}{(n - r)!} & = & \frac{n!}{(n - (r + 1))!} & [∵ {}^{n}P_{r} = \frac{n!}{(n - r)!}] \\ \Rightarrow & \frac{1}{(n - r)!} & = & \frac{1}{(n - r - 1)!} \\ \Rightarrow & \frac{1}{(n - r) (n - r - 1)!} & = & \frac{1}{(n - r - 1)!} \\ \Rightarrow & n - r & = & 1 \\ \Rightarrow & n & = & r + 1 & . . . [5] \end{array}$
Step 2: Solve for the required value
Comparing equations $[4]$ and $[5]$ , we get,
$\begin{array}{crcl} 2 r - 1 & = & r + 1 \\ \Rightarrow & r & = & 2 \end{array}$
Substituting the value of $r$ in equation $[5]$
$\begin{array}{crcl} ∴ & n & = & r + 1 \\ \Rightarrow & n & = & 2 + 1 \\ \Rightarrow & n & = & 3 \end{array}$
Hence the correct option is option(A) i.e. $3$

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