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

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2^{19} +

^{20} C_{10}

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n o n e o f t h e s e

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Solution

The correct option is A $n o n e o f t h e s e$
$^{20} C_{0} +^{20} C_{1} +^{20} C_{2} +^{20} C_{10}$
$= 1 + 20 + \frac{20 (19)}{2} + \frac{20!}{10! (10!)}$
$= 21 + 190 + \frac{20!}{(10!)^{2}}$
$= 211 + \frac{20!}{(10!)^{2}}$
Hence the answer is none of these.

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