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Function for population size from known frequencies?

Richard Frost, Frost Concepts
Posted 2 years ago

Perhaps there is a built-in function for this?

I have vectors of integers. All vectors have the same number of elements c. There are c values in the domain. The values in each vector need not be unique but there may not be duplicate vectors.

Each set of these vectors has a known distribution. For example, for c=5 there is a set with frequencies {1,2,2}. One member of that set looks like {a,b,b,c,c}.

The function I'm seeking would calculate the size (number of vectors) with F[5,{1,2,2}].

In lieu of a function I've attempted to define a formula. There appears to be two trivial distributions: {c} and {1, 1, …, 1}. The former has size c and the latter c!. The others involve products of binomial coefficients.
POSTED BY: Richard Frost
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Jim Baldwin
Jim Baldwin
Posted 2 years ago

I've added in a formula for the total number of arrangements along with a rationale for constructing that formula.
POSTED BY: Jim Baldwin
Jim Baldwin
Jim Baldwin
Posted 2 years ago
POSTED BY: Jim Baldwin

Hi Jim, As an example, consider the general population of vectors of length n = 5 composed of n letters: a,b,c,d,e. This population has well known size n^n.

I'm interested in subsets of this population defined by frequencies of the letters (in this example). Here's an example set of frequencies: {1,2,2}. It has members that look like {a,b,b,c,c}, {a,b,c,b,c}, ..., {d,e,d,e,a}.

Is there a Mathematica function that would calculate the size of this subset, given 5 letters and frequences {1,2,2} ?
POSTED BY: Richard Frost
Jim Baldwin
Jim Baldwin
Posted 2 years ago

Are you looking for Multinomial ?

Multinomial[1, 2, 2]
(* 30 *)

But Multinomial[c] equals 1 for any integer value of c. You've show one member of a set. If you could show a few other members (or all members) that would help determine how to obtain the count of members in general.

As an example:

Multinomial[1, 3, 4]
(* 280 *)
Length[Permutations[{a, b, b, b, c, c, c, c}]]
(* 280 *)
POSTED BY: Jim Baldwin
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