First slide

Combinations

Question

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

Easy

A

$75$
B

$74$
C

$67$
D

$45$

Solution

$\begin{array}{l} We know {}^{n}C_{r} is greatest at middle term(s) \\ a = {}^{19}C_{p} = {}^{19}C_{10} = {}^{19}C_{9}, b = {}^{20}C_{q} = {}^{20}C_{10}, c = {}^{21}C_{6} = {}^{21}C_{10} = {}^{21}C_{11} \\ Now, using {}^{n}{C_{r}} = \frac{n}{r} ({}^{n - 1}{C_{r - 1}}), we get b=2 ({}^{19}{C_{9}}), c = (\frac{21}{11} × \frac{20}{10} × {}^{19}{C_{9}}) \\ \Rightarrow a + b + c = {{}^{19}C}_{9} (1 + 2 + \frac{42}{11}) = {}^{19}{C_{9}} (\frac{75}{11}) \\ \frac{11}{({}^{19}{C_{9}})} (a + b + c) = 75 \end{array}$

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