if ^{n}C_{r} - 1 = 36,^{n}C_{r} = 84 and ^{n}C_{r+1 }= 126, then find^{ r}C_{2}.

[**HINT** : form eqn using ^{n}Cr/^{n}C_{r+1} and ^{n}C_{r}/^{n}C_{r-1} to find the value of r.]

^{n}C_{ r }/ ^{n}C_{r+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)

^{n}C_{r}/^{n}C_{r-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, ^{r}C_{2 }= ^{3}C_{2 }= 3