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The number is 5! = 120

Q: What is the possible number of symmetric relations on a set of 5 elements?

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2 power 20

In a skew symmetric matrix of nxn we have n(n-1)/2 arbitrary elements. Number of arbitrary element is equal to the dimension. For proof, use the standard basis.Thus, the answer is 3x2/2=3 .

true

5 expressed as a Roman numeral is V which is symmetrical

A circle is symmetric about ANY diameter. The number of possible diameters of the same circle is infinite. And on the same principle, a sphere will have lines of symmetry in every direction in 3 dimensions.

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2^(n^2+n)/2 is the number of symmetric relations on a set of n elements.

the total no of reflexive relation on an n- element set is 2^(n^2-n).

2 power 20

make a table as I did below for the set {a,b,c} with 3 elements. A table with all n elements will represent all the possible relations on that set of n elements. We can use the table to find all types of relations, transitive, symmetric etc. | a | b | c | --+---+---+---+ a | * | | | b | | * | | c | | | * | The total number of relations is 2^(n^2) because for each a or b we can include or not include it so there are 2 possibilities and there are n^2 elements so 2^(n^2) total relations. A relation is reflexive if contains all pairs of the form {x,x) for any x in the set. So this is the diagonal of your box. THESE ARE FIXED! No, in reflexive relation we still can decide to include or not include any of the other elements. So we have n diagonal elements that are fixed and we subtract that from n^2 so we have 2^(n^2-n) If you do the same thing for symmetric relations you will get 2^(n(n+1)/2). We get this by picking all the squares on the diagonal and all the ones above it too.

It tells you the atomic number, the number of protons, the density, its relations to other elements, and its characteristics.

In a skew symmetric matrix of nxn we have n(n-1)/2 arbitrary elements. Number of arbitrary element is equal to the dimension. For proof, use the standard basis.Thus, the answer is 3x2/2=3 .

Yes.

yes

90 naturally occurring elements, unknown number possible elements.

no that's not possible

It depends what you mean by symmetric. If you mean that there are the same number of factors on each branch, then the answer is No. eg 20 / \ 4 5 / \ 2 2

Two primes are symmetric primes of a natural number (n) if their average is n. For example 10, has two pairs of symmetric primes. 7 and 13, and 3 and 17 because their averages are 10.