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    total number of different partitions of a set having four elements is

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    How many different partitions with exactly two parts can be made of the set {1,2,3,4}? There are 4 elements in this list that need to be partitioned into 2 parts. I wrote these out and got a total ...

    How many different partitions with exactly n parts can be made of a set with k-elements?

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    How many different partitions with exactly two parts can be made of the set {1,2,3,4}? There are 4 elements in this list that need to be partitioned into 2 parts. I wrote these out and got a total of 7 different possibilities:

    {{1},{2,3,4}} {{2},{1,3,4}} {{3},{1,2,4}} {{4},{1,2,3}} {{1,2},{3,4}} {{1,3},{2,4}} {{1,4},{2,3}}

    Now I must answer the same question for the set {1,2,3,...,100}. There are 100 elements in this list that need to be partitioned into 2 parts. I know the largest size a part of the partition can be is 50 (that's 100/2) and the smallest is 1 (so one part has 1 number and the other part has 99). How can I determine how many different possibilities there are for partitions of two parts without writing out extraneous lists of every possible combination? Can the answer be simplified into a factorial (such as 12!)?

    Is there a general formula one can use to find how many different partitions with exactly n parts can be made of a set with k-elements?

    mathdiscrete-mathematicsdata-partitioning

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    edited Sep 3, 2012 at 17:18

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    asked Feb 16, 2012 at 17:54

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    2 Answers

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    1) stackoverflow is about programming. Your question belongs to https://math.stackexchange.com/ realm.

    2) There are 2n subsets of a set of n elements (because each of n elements may either be or be not contained in the specific subset). This gives us 2n-1 different partitions of a n-element set into the two subsets. One of these partitions is the trivial one (with the one part being an empty subset and other part being the entire original set), and from your example it seems you don't want to count the trivial partition. So the answer is 2n-1-1 (which gives 23-1=7 for n=4).

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    answered Feb 16, 2012 at 18:05

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    penartur

    Feb 17, 2012 at 9:41

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    The general answer for n parts and k elements would be the Stirling number of the second kind S(k,n).

    Please beware that the usual convention is with n the total number of elements, thus S(n,k)

    Computing the general formula is quite ugly, but doable for k=2 (with the common notation) :

    Thus S(n,2) = 1/2 ( (+1) * 1 * 0n +(-1) * 2 * 1n + (+1) * 1 * 2n ) = (0-2+2n)/2 = 2n-1-1

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    edited Feb 8, 2017 at 14:58

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    answered Dec 1, 2014 at 22:41

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    Not the answer you're looking for? Browse other questions tagged mathdiscrete-mathematicsdata-partitioning or ask your own question.

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    स्रोत : stackoverflow.com

    How to find all partitions of a 4

    Answer: As one of the comments suggested, you can use the Stirling numbers of the second kind - Wikipedia, S(n,k), to calculate the number of ways to separate n objects into k partitions. So S(4,1)=1 is the number of ways to put 4 objects into 1 partition, S(4,2)=7 is the number of ways to have 2...

    How do I find all partitions of a 4-element set?

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    Sort Glenn Redd

    B.S. in Mathematics, Kennesaw State University (Graduated 2016)Author has 1.9K answers and 2.3M answer views4y

    As one of the comments suggested, you can use the Stirling numbers of the second kind - Wikipedia

    , S(n,k), to calculate the number of ways to separate n objects into k partitions. So S(4,1)=1 is the number of ways to put 4 objects into 1 partition, S(4,2)=7 is the number of ways to have 2 partitions, S(4,3)=6 is 3 partitions, and S(4,4)=1 is 4 partitions. So the sum of these 1+7+6+1=15 is the number of total possible partitions of a 4 element set.

    This is also the 5th Bell number (since the 1st bell number is the number of partitions for 0 elements, that means 4 elements is the 5th), but one of the ways to calculate the Bell number is to add up all of the Stirling numbers of the second kind for the given number of elements, so…

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    What is a partition of a set?

    > What is a partition of a set?

    A Partition of a set [ https://en.m.wikipedia.org/wiki/Partition_of_a_set ],

    A A

    , is a set of subsets,

    B i Bi , of A A such that: ⋃ i B i =A ⋃iBi=A and ∀i≠j: B i ∩ B j =∅ ∀i≠j:Bi∩Bj=∅

    That is it is a splitting of the set into disjoint subsets that cover the original set. Every member of the original set is in one and only one subset.

    You should be familiar with a partition from your experiences at school. Your year (the set of people your age at your school) was probably partitioned into classes. You were in one...

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    How many subsets does a set with 4 elements have?

    2^4 = 16. The empty set, {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}, and {A, B, C, D} itself.

    Generally, to construct a subset, list all elements of the set and to each element assign either YES (belongs to the subset) or NO (does not belong to the subset). This can be done in 2 ways for each element; therefore, if the original set has n elements, the total number of possible choices is 2*2*2*…*2 (n times), i.e. 2^n.

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    How is it possible to have a set with one element only?

    Consider for example the set of all planets you've been to. My guess is this set has a single element.

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    Why is the empty set considered a set?

    My little brother had trouble with the concept of empty sets when he was first learning about them in high school. This is how I explained it to him (this might not be entirely mathematical, kindly correct me if anything I say is not logical):

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    Can an element be a set?

    Depending on your set theory, it’s possible that it can’t be anything else.

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    Sets can always be elements. The complicated question is whether an element can be something other than a set.

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