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Permutations and Combinations question

2021 · 25 Jul · Shift 2 · Q39
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Permutations and Combinations question

2021 · 25 Jul · Shift 2 · Q39

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
If nPr=nPr+1{}^n{P_r} = {}^n{P_{r + 1}}nPr​=nPr+1​ and nCr=nCr−1{}^n{C_r} = {}^n{C_{r - 1}}nCr​=nCr−1​, then the value of r is equal to :
  1. A
    1
  2. B
    4
  3. C
    2
  4. D
    3
View written solutionFree

Correct answer: C

  1. We use the formulas

nPr=n!(n−r)!,nCr=n!r!(n−r)!^{n}P_r = \frac{n!}{(n-r)!}, \qquad {}^{n}C_r = \frac{n!}{r!(n-r)!}nPr​=(n−r)!n!​,nCr​=r!(n−r)!n!​

  1. Given

nPr=nPr+1^{n}P_r = {}^{n}P_{r+1}nPr​=nPr+1​

Substitute the formula:

n!(n−r)!=n!(n−r−1)!\frac{n!}{(n-r)!} = \frac{n!}{(n-r-1)!}(n−r)!n!​=(n−r−1)!n!​

Cancel n!n!n!:

1(n−r)!=1(n−r−1)!\frac{1}{(n-r)!} = \frac{1}{(n-r-1)!}(n−r)!1​=(n−r−1)!1​

So,

(n−r)!=(n−r−1)!(n-r)! = (n-r-1)!(n−r)!=(n−r−1)!

Using

(n−r)!=(n−r)(n−r−1)!(n-r)! = (n-r)(n-r-1)!(n−r)!=(n−r)(n−r−1)!

we get

(n−r)(n−r−1)!=(n−r−1)!(n-r)(n-r-1)! = (n-r-1)!(n−r)(n−r−1)!=(n−r−1)!

Hence,

n−r=1  ⟹  n=r+1n-r = 1 \implies n = r+1n−r=1⟹n=r+1

  1. Now use the second condition:

nCr=nCr−1^{n}C_r = {}^{n}C_{r-1}nCr​=nCr−1​

Substitute the formula:

n!r!(n−r)!=n!(r−1)!(n−r+1)!\frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!}r!(n−r)!n!​=(r−1)!(n−r+1)!n!​

A standard identity says that

nCr=nCr−1  ⟹  r=n+12^{n}C_r = {}^{n}C_{r-1} \implies r = \frac{n+1}{2}nCr​=nCr−1​⟹r=2n+1​

Now from n=r+1n=r+1n=r+1,

r=(r+1)+12=r+22r = \frac{(r+1)+1}{2} = \frac{r+2}{2}r=2(r+1)+1​=2r+2​

So,

2r=r+2  ⟹  r=22r = r+2 \implies r=22r=r+2⟹r=2

  1. Check with options:
  • A: 111
  • B: 444
  • C: 222 ✅
  • D: 333

Therefore, the correct answer is

2\boxed{2}2​

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