Verified Solution • 2025–26 Session moderate

QJEEM21Ma0dab: JEE Main 2021 Mathematics - Class 11 JEEM21Ma0dab Solution: If and , then the value of r is equal to :

If nPr=nPr+1{}^n{P_r} = {}^n{P_{r + 1}} and nCr=nCr−1{}^n{C_r} = {}^n{C_{r - 1}} , then the value of r is equal to :

Multiple Choice Options

  • A 1
  • B 4
  • C 2
  • D 3

Step-by-Step Verified Solution

nPr=nPr+1⇒n!(n−r)!=n!(n−r−1)!{}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}

⇒(n−r)=1\Rightarrow (n - r) = 1 .....(1)

nCr=nCr−1{}^n{C_r} = {}^n{C_{r - 1}}

⇒n!r!(n−r)!=n!(r−1)!(n−r+1)!\Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}

⇒1r(n−r)!=1(n−r+1)(n−r)!\Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}

⇒n−r+1=r\Rightarrow n - r + 1 = r

⇒n+1=2r\Rightarrow n + 1 = 2r ..... (2)

From (1) and (2), 2r−1−r=1⇒r=22r - 1 - r = 1 \Rightarrow r = 2

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