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I have four letters, $A B C D$ , and I need all possible combinations when the order doesn't matter.

I thought this should be $2^k$ , with $ k$ being $4 $ .

But I can only count $15$, is there a reason for this? Am I missing one?

A
B
C
D
A,B
A,C
A,D
B,C
B,D
C,D
A,B,C
A,B,D
A,C,D
B,C,D
A,B,C,D
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4
  • 4
    $\begingroup$ case with no letters $\endgroup$
    – Fomalhaut
    Sep 16, 2019 at 8:21
  • 2
    $\begingroup$ Don't you mean "I can only count 15" ??? $\endgroup$
    – user65203
    Sep 16, 2019 at 8:23
  • $\begingroup$ yes thanks!!!!!!!!! $\endgroup$ Sep 16, 2019 at 8:26
  • $\begingroup$ You count 15 not 16.... $\endgroup$
    – dmtri
    Sep 16, 2019 at 8:27

2 Answers 2

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You are missing the combination containing no letters at all. A set of $k$ elements has $2^k$ subsets, but only $2^k - 1$ non-empty subsets.

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$2^k$ is due to the fact that you can take or not take the letters independently, so every binary choices multiply with each other. One of the combinations is "take none".

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