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All choices of two different numbers from this set give rise to different sums apart from \(1 + 33\) and \(2 + 32\).

The number of ways to choose \(2\) numbers from the set {\(1, 2, 4, 8, 16, 32, 33\)} of \(7\) numbers is

7C

2 = \(\frac{(7*6)}{(1*2)}\) = \(21\).

Since two of these choices, \(1, 33\) and \(2, 32\), have the same sum, we need to subtract \(1\) from \(21\).

Then we have \(21 – 1 = 20\) possible choices.

Therefore, the answer is E.

Answer: E

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