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If $^{n} C_{r - 1} = 36,^{n} C_{r} = 84 {and}^{n} C_{r + 1} = 126,$ then the value of $r$ is equal to

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Correct option is C

We have nCr−1 nCr=3684⇒n!(r−1)!(n−r+1)!r!(n−r)!n!=37⇒ rn−r+1=37 ⇒ 10r=3n+3Similarly nCr nCr+1=84126 ⇒ r+1n−r=23⇒ 5r+3=2n ⇒ 5r+3=2nSolving we obtain r = 3.

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If nCr−1=36,nCr=84 and nCr+1=126, then the value of r is equal to

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If $^{n} C_{r - 1} = 36,^{n} C_{r} = 84 {and}^{n} C_{r + 1} = 126,$ then the value of $r$ is equal to