lf a, b and c are the greatest values of  19Cp′20Cq and 21Cr respectively, the

Since a = maximum of  19Cp=19C9=19C10-------(i)         b = maximum of  20Cq=20C10 ------(ii)and  c = maximum of  21Cr=21C10=21C11-------(iii)then  ab= 19C10 20C10=19!10!9!20!=1020=12⇒a1=b2and bc= 20C10 21C11=20!10!10!21!11!10!=1121⇒b11=c21a11=b22=c42Hence, option (d) is correct.