I'm solving a problem that uses Combinations and it's driving me crazy. The question requires the number of unique subsets (order does not matter) from 5 elements. I believe the answer should just be C(5, 3) correct?

$\frac{5!}{3!(5-3)!}$ that is.

However, as I manually list out all these sets. I am only getting 9, as my answer, not 10.... C(5, 3) = 10.

$(A, B, C, D, E)\\ (A, B, C)\\ (A, B, D)\\ (A, B, E)\\ (A, C, D)\\ (A, C, E)\\ (A, D, E)\\ (B, C, D)\\ (B, D, E)\\ (C, D, E)\\ $

The question requires that there cannot be any duplicate element sets, so the order does not matter. Hence the set (A, B, C) = (B, A, C). Is this not a combination or am I just missing one of the subsets? Thanks.

  • 4
    $\begingroup$ missing $(B,C,E)$... $\endgroup$ – lulu Jan 24 '18 at 2:56
  • 1
    $\begingroup$ Man, I'm an idiot and you're a live saver. So it was a combination afterall correct? $\endgroup$ – Wallace Jan 24 '18 at 2:58
  • $\begingroup$ It was correct, well done for that. $\endgroup$ – астон вілла олоф мэллбэрг Jan 24 '18 at 3:05

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