If Prn=Pr+1n and Crn=Cr-1n, then the value of r is equal to:

Correct answer: (C) 2. We have, Prn=Pr+1n ⇒(n!)/(n-r!)=(n!)/(n-r-1!) ⇒n-r=1 …1 and Crn=Cr-1n ⇒(n!)/(r!n-r!)=(n!)/(r-1!n-r+1!) ⇒1/(rn-r!)=1/(n-r+1n-r!) ⇒n-r+1=r ⇒n+1=2r …2 From equation 1 & 2, we get 2r-1-r=1 ⇒r=2.

Q48 / 90 · JEE Main 2021 (25 Jul Shift 2) · Mathematics

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

A
$1$
B
$4$
Correct answer
C
$2$
D
$3$

Solution

We have, ${}^{n}P_{r} = {}^{n}P_{r + 1}$

$\Rightarrow \frac{n!}{(n - r)!} = \frac{n!}{(n - r - 1)!}$

$\Rightarrow (n - r) = 1 \dots (1)$

and ${}^{n}C_{r} = {}^{n}C_{r - 1}$

$\Rightarrow \frac{n!}{r! (n - r)!} = \frac{n!}{(r - 1)! (n - r + 1)!}$

$\Rightarrow \frac{1}{r (n - r)!} = \frac{1}{(n - r + 1) (n - r)!}$

$\Rightarrow n - r + 1 = r$

$\Rightarrow n + 1 = 2 r \dots (2)$

From equation $(1) & (2),$ we get

$2 r - 1 - r = 1$

$\Rightarrow r = 2 .$

Hence, the correct answer is option: 'C'.

If Prn=Pr+1n and Crn=Cr-1n, then the value of r is equal to:

Solution

Relevant Questions

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