Question
The sum $$^{20}{C_0} + {\,^{20}}{C_1} + {\,^{20}}{C_2} + ..... + {\,^{20}}{C_{10}}$$ is equal to
A.
$${2^{20}} + \frac{{20!}}{{{{\left( {10!} \right)}^2}}}$$
B.
$${2^{19}} - \frac{1}{2} \cdot \frac{{20!}}{{{{\left( {10!} \right)}^2}}}$$
C.
$${2^{19}} + {\,^{20}}{C_{10}}$$
D.
None of these
Answer :
None of these
Solution :
Let $$S = {\,^{20}}{C_0} + {\,^{20}}{C_1} + {\,^{20}}{C_2} + ..... + {\,^{20}}{C_{10}}$$
$$S = {\,^{20}}{C_{20}} + {\,^{20}}{C_{19}} + {\,^{20}}{C_{18}} + ..... + {\,^{20}}{C_{10}}\,\,\,\left( {\because \,{\,^n}{C_r} = {\,^n}{C_{n - r}}} \right)$$
Adding, $$2S = \left( {^{20}{C_0} + {\,^{20}}{C_1} + ..... + {\,^{20}}{C_{20}}} \right) + {\,^{20}}{C_{10}} = {2^{20}} + {\,^{20}}{C_{10}}.$$