If a, b and c are the greatest values of Cp   19 , 20Cq and  Cr   21  respectively, then:

We know  Cn2   n(if n is even) and Cn-12   nor Cn+12   n (if n is odd) are greatest at middle term(s)∴a=Cp   19=C10   19=C9   19and b=Cq   20=C10   20and c=Cr   21=C10   21=C11   21Now, using Cr   nCr−1   n−1=nr , we get aC9   19=b2010.C9   19=c2111.2010C9   19⇒a1=b2=c4211⇒a11=b22=c42