User:IssaRice/Summary of counting techniques: Difference between revisions

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| || <math>\{\{a_1, \ldots, a_k\} : a_1,\ldots, a_k \in A\}</math> || <math>\sum_{i=1}^k \binom n i</math>
| || <math>\{\{a_1, \ldots, a_k\} : a_1,\ldots, a_k \in A\}</math> || <math>\sum_{i=1}^k \binom n i</math>
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| || <math display="inline">\{ f : A \to \mathbf N \mid \sum_{a \in A} f(a) = k\}</math>
| || <math display="inline">\{ f : A \to \mathbf N \mid \sum_{a \in A} f(a) = k\}</math> (multisets with cardinality <math>k</math>) ||
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| || <math>\{\{a_1, \ldots, a_n\} : a_1,\ldots, a_n \in A\}</math> || <math>\sum_{i=1}^n \binom n i = 2^n - 1</math> (a quick way to see this identity is that we want the power set without the empty set)
| || <math>\{\{a_1, \ldots, a_n\} : a_1,\ldots, a_n \in A\}</math> || <math>\sum_{i=1}^n \binom n i = 2^n - 1</math> (a quick way to see this identity is that we want the power set without the empty set)

Latest revision as of 02:38, 14 August 2019

Let be a set with elements, and let be a set with elements.

Description Set representing counting problem number of ways to count
Pick things from with replacement
(multisets with cardinality )
(a quick way to see this identity is that we want the power set without the empty set)