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$If a, b and c are the greatest values of {}^{19}C_{p}, {}^{20}C_{q} and {}^{21}C_{r} respectively, then:$ $\frac{11}{({}^{19}{C_{9}})} (a + b + c) =$

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Correct option is A

We know Cr n is greatest at middle term(s) a=Cp 19=C10 19=C9 19, b=Cq 20=C10 20, c=C6 21=C10 21=C11 21 Now, using Cr n=nr Cr-1 n-1, we get b=2 C9 19, c =2111×2010×C9 19 ⇒a+b+c =C 1991+2+4211=C9 19751111C9 19a+b+c=75

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If a, b and c are the greatest values of Cp 19,Cq 20 and Cr 21 respectively, then: 11C9 19a+b+c=

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$If a, b and c are the greatest values of {}^{19}C_{p}, {}^{20}C_{q} and {}^{21}C_{r} respectively, then:$ $\frac{11}{({}^{19}{C_{9}})} (a + b + c) =$