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

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a10=b11=c42

a11=b22=c21

a10=b11=c21

a11=b22=c42

answer is D.

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Detailed Solution

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.

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