MCQ
$^{10}{C_1}{ + ^{10}}{C_3}{ + ^{10}}{C_5}{ + ^{10}}{C_7}{ + ^{10}}{C_9} = $
- ✓${2^9}$
- B${2^{10}}$
- C${2^{10}} - 1$
- DNone of these
${2^{n - 1}} = {\,^n}{C_0} + {\,^n}{C_2} + {\,^n}{C_4} + .... = {\,^n}{C_1} + {\,^n}{C_3} + {\,^n}{C_5} + .....$
So, $^{10}{C_1} + {\,^{10}}{C_3} + {\,^{10}}{C_5} + ..... + {\,^{10}}{C_9} = {2^{10 - 1}} = {2^9}$
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