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visits member for 3 years, 9 months
seen Jun 15 '12 at 23:08

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awarded  Enthusiast
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awarded  Teacher
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comment Why is the number of possible subsequences $2^n$?
I mean, in this case in which you don't consider the order of the elements in each subsequence, the proof Manoj R provides suffice. Otherwise, for each subset of length k you also consider the number of its permutations: so you have C(n,0)*0! + C(n,1)*1! + C(n,2)*2! + ... + C(n,n)*n!.
Feb
24
comment Why is the number of possible subsequences $2^n$?
@Manish: i think what Qiaochu mean is that in this case to involve binomial theorem is just... too much:) I think the proof you propose in your comment becomes useful (and meaningful) if you consider also the order of the elements in each subsequence. Which is not the case.
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awarded  Editor
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revised Why is the number of possible subsequences $2^n$?
added 131 characters in body
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24
answered Why is the number of possible subsequences $2^n$?
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awarded  Scholar
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30
accepted $(\{0, 1, … , 2^n-1\}, \oplus)$ groups
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awarded  Student
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asked $(\{0, 1, … , 2^n-1\}, \oplus)$ groups