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How many subsets of three different integers between $1$ and $90$ inclusive are there whose sum is an even number?

If I am not wrong then there are $45$ odd numbers and $45$ even . Also $$\text{even}+\text{even}+\text{even}=\text{even},\quad \text{odd}+\text{odd}+\text{even}=\text{even}.$$

But we need 3 different integers subset. How should I proceed?

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    $\begingroup$ odd +even+even $\neq$ even $\endgroup$ – Jaideep Khare Nov 21 '17 at 7:01
  • $\begingroup$ Thanks .I mistakenly did $\endgroup$ – Kavita Sahu Nov 21 '17 at 7:02
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The sum of three numbers is even iff they are all even or there is one even and two odds. Hence if the three numbers belong to a set with $E$ even numbers and $O$ odd numbers then, in the first case the number of ways is $\binom{E}{3}$ and in the second case $\binom{E}{1}\cdot \binom{O}{2}$.

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Hint for an easier calculation:

You might find it useful to consider the sets $\{a,b,c\}$ and $\{(91-a), (91-b), (91-c)\}$ and show that half the sets of three distinct elements have an even sum.

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$$\binom{45}{2}\binom{45}{1}+\binom{45}{3}.$$

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