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

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.