if nCr - 1 = 36,nCr = 84 and nCr+1 = 126, then find rC2.

[HINT : form eqn using nCr/nCr+1 and nCr/nCr-1 to find the value of r.]

nC r / nCr+1 = r+1 / n-r =84/126 =2/3

r+1 / n-r =2/3 , 3r+3=2n-2r , ( 2n-5r-3=0 )x 3 = 6n-15r-9=0 .......(i)

nCr/nCr-1 =n-r+1 / r =84/36 =7/3

n-r+1 / r = 7/3 ,,, 3n-3r+3=7r, (3n-10r+3=0) x2 = 6n-20r+6=0.......(II)

from i - ii , we have r=3

therefore, rC2 = 3C2 = 3

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From eq 1 and 2, 3n+13=4n+4 N=9 & 3(9)+13=10r R=4

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