The sum $^{20} C_{0} +^{20} C_{1} +^{20} C_{2} + . . . . . +^{20} C_{10}$ is equal to

2^{20} + \frac{20!}{(10!)^{2}}

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2^{19} + \frac{1}{2} . \frac{20!}{(10!)^{2}}

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2^{19} +^{20} C_{10}

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none of these

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Solution

The correct option is C $2^{19} + \frac{1}{2} . \frac{20!}{(10!)^{2}}$
Let
$S =^{20} C_{0} +^{20} C_{1} +^{20} C_{2} + . . . . . . . +^{20} C_{10}$
$\Rightarrow 2 S = (^{20} C_{0} +^{20} C_{0}) + (^{20} C_{1} +^{20} C_{1}) + (^{20} C_{2} +^{20} C_{2}) + . . . . . . . + (^{20} C_{9} +^{20} C_{9}) + (^{20} C_{10} +^{20} C_{10})$
$\Rightarrow 2 S -^{20} C_{10} = (^{20} C_{0} +^{20} C_{20}) + (^{20} C_{1} +^{20} C_{19}) + (^{20} C_{2} +^{20} C_{18}) + . . . . . . . + (^{20} C_{9} +^{20} C_{11}) +^{20} C_{10}$
(Using $^{n} C_{r} =^{n} C_{n - r}$ )
$\Rightarrow 2 S -^{20} C_{10} =^{20} C_{0} +^{20} C_{1} +^{20} C_{2} + . . . . . . . +^{20} C_{19} +^{20} C_{20}$
$\Rightarrow 2 S -^{20} C_{10} = 2^{20}$ (Using $^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . . . . . +^{n} C_{n} = 2^{n}$ )
$\Rightarrow S = \frac{2^{20} +^{20} C_{10}}{2}$
$\Rightarrow S = 2^{19} + \frac{1}{2} . \frac{20!}{(10!)^{2}}$

Hence, Option B is correct.

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