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 \text{ and all }a_i\text{ distinct}\}</math> || <math>P(n,k) = \frac{n!}{(n-k)!} = n(n-1)\cdots (n-(k+1))</math>
| || <math>\{(a_1, \ldots, a_k) : a_1,\ldots, a_k \in A \text{ and all }a_i\text{ distinct}\}</math> || <math>P(n,k) = \frac{n!}{(n-k)!} = n(n-1)\cdots (n-(k+1))</math>
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| || <math>\{(a_1, \ldots, a_n) : a_1,\ldots, a_n \in A \text{ and all }a_i\text{ distinct}\}</math> || <math>P(n,n) = n!</math>
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| || <math>\{\{a_1, \ldots, a_k\} : a_1,\ldots, a_k \in A \text{ and all }a_i\text{ distinct}\}</math> || <math>\binom n k = P(n,k)/(k!) = \frac{n!}{k!(n-k)!}</math>
| || <math>\{\{a_1, \ldots, a_k\} : a_1,\ldots, a_k \in A \text{ and all }a_i\text{ distinct}\}</math> || <math>\binom n k = P(n,k)/(k!) = \frac{n!}{k!(n-k)!}</math>

Revision as of 02:24, 14 August 2019

Let A be a set with n elements, and let B be a set with m elements.

Description Set representing counting problem number of ways to count
{(a1,,ak):a1,,akA} nk
{{a1,,ak}:a1,,akA} i=1k(ni)
{{a1,,an}:a1,,anA} i=1n(ni)=2n1
{(a1,,ak):a1,,akA and all ai distinct} P(n,k)=n!(nk)!=n(n1)(n(k+1))
{(a1,,an):a1,,anA and all ai distinct} P(n,n)=n!
{{a1,,ak}:a1,,akA and all ai distinct} (nk)=P(n,k)/(k!)=n!k!(nk)!
{(a,b):aA and bB} nm