The number of ways one can choose a set of distinct positive integers, each smaller than or equal to $20,$ such that there sum is odd

2^{18}

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2^{19}

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2^{20}

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2^{21}

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Solution

The correct option is B $2^{19}$
Either the sum will be even or odd
choosing
$0 number =^{20} C_{0} 1 number =^{20} C_{1} 2 number =^{20} C_{2} . . . 20 number =^{20} C_{20}$
$∴ odd sum = \frac{^{20} C_{0} +^{20} C_{1} +^{20} C_{2} + . . . . . +^{20} C_{20}}{2}$
$= \frac{2^{20}}{2} = 2^{19}$

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