Enumerations for Compositions and Complete Homogeneous Symmetric Polynomial

We count the number of occurrences of t as the summands (i) in the compositions of a positive integer n into r parts; and (ii) in all compositions of n; and subsequently obtain other results involving compositions. The initial counting further helps to solve the enumeration problems for complete homogeneous symmetric polynomial.

Evidently ℎ  ( 1 , … ,   ) has some enumerating problems.We give the solutions of the problems from some combinatorial enumerations for the compositions of a positive integer.The paper has two main parts: (a) counting for compositions and (b) counting for ℎ  ( 1 , … ,   ).
The main results are as shown: 1.The number of occurrences of an integer t as the summands in the compositions of n into r parts = ( -1  2 ) , , -+ 2. The number of occurrences of t as the summands in all compositions of n = ( -+ 3) -2 , > 3. The number of occurrences of t as the exponents in different terms among all the terms of ℎ  ( 1 , … ,   ) 4. The number of occurrences of each variable as the bases in different terms among all the terms of ℎ  ( 1 , … ,   )

Counting for Compositions
For counting, we use some simple notations as shown.
1. Compositions of n into r summands = C(n, r).
The notation C(n, r) without any qualification means all compositions of n into r summands.Otherwise we use an adjective to specify the compositions.For example, we write simply 'some C(n, r)' to mean some compositions of n into r summands.

Number of occurrences of
4. Some particular C(n, r) that start with a common summand k = k + C(nk, r -1).
We use the symbol of equivalence (≡) between C(n, r) and its implication; and similarly between k + C(nk, r -1) and its implication.

Number of Occurrences of t in the Compositions of n into r Parts
The number of the compositions of a positive integer n into r parts or summands is ( 1  1 ).This is a known result.Here we obtain the result in a process of recursive substitution starting with a basic sequential arrangement of n into r summands.The procedure and result lead to count the number of occurrences of t in the compositions of n into r parts.
First we count NC(n, r) for n ≥ r ≥ 1.
(a) By convention, n itself is a composition of n so that r is equal to 1 for the composition.Therefore, for n ≥ 1 and r = 1, we have: NC(n, r) = NC(n, 1) = 1.

(b)
For n ≥ r ≥ 2, we can write a basic sequential arrangement of C(n, r) in the following way.
In general for n ≥ r ≥ 2, NC(n, r) = ( 1  1 ); and including the initial result with this, we get: We count below N(t)C(n, r) by the above process and results.
The basic conditions of n, t and r are: n ≥ t ≥ 1, n ≥ r ≥ 1 and r ⋛ t.
(a) When n ≥ 1 and r = 1 then t = n so that N(t)C(n, r) = N(n)C(n, 1) = 1.For n > t ≥ 1, mathematically we can write: N(t)C(n, 1) = 0.These are the initial results in the counting of N(t)C(n, r).
(b) When n and r are the fixed integers for n ≥ r ≥ 2 then we find some fixed C(n, r).In one of these C(n, r), each of r -1 summands is smallest or 1 so that the rest is greatest.Consequently the greatest value of a summand t is nr + 1.That is, when n ≥ r ≥ 2 then the condition of t is: nr + 1 ≥ t ≥ 1.From (1.1) we find: (i) Some C(n, r) start with a common summand t; and these are: t + C(nt, r -1).
(ii) t can occur at other places of different compositions under some or all of The smallest positive integer: 1 can occur as the summands in different compositions under ).Yet occurrences of t ≥ 2 have some limitations.Since nr + 1 ≥ t, it follows that if nr ≤ t -2 then t cannot occur in a C(n, r).More precisely, we cannot find the occurrences of 2 in C(r -1, r -1); 3 in C(r, r -1) and C(r -1, r -1); In general t cannot occur in any composition under C(r Then from (i) and (ii), we get: for n ≥ r ≥ 2 and nr + 1 ≥ t ≥ 1, When r = 2, then for n ≥ 2 and n -1 ≥ t ≥ 1, Evidently N(t)C(n, 2) is constant and independent of t and n. (3.1) is the primary case of (3); and then the rest of (3) is: for n ≥ r ≥ 3 and nr + 1 ≥ t ≥ 1, 2) Now our aim is to count N(t)C(n, r) applying (3.1) and (3.2).

To count N(t)C(n, r) for t = 1
For n ≥ 2, N( 1 In general for n ≥ r ≥ 2, N(1)C(n, r) = (2  2 ) 2. To count N(t)C(n, r) for t = 2 n, r and t have the conditions: n ≥ r ≥ 2 and nr + 1 ≥ t ≥ 1. Hence when t = 2 then the conditions of n and r are: n -1 ≥ r ≥ 2 and n ≥ 3.
. By the similar operation, we get:

Number of Occurrences of t in All Compositions of n
Let the number be denoted by N(t)C(n).

Number of Terms of the Polynomial
The result is known.Here we count the number applying (2) and Vandermonde's identity.Let some terms of the polynomial contain some fixed m of k variables.The number of these terms = NC(r, m) = (  1 1 ).We have k ⋛ r in the problem.Hence we find: Case 2: When 1 ≤ m ≤ r ≤ k then the number of terms It follows that the number of terms does not depend on equality or any inequality between k and r, which are all taken into consideration in the process of solution.Thus we find: The number of terms of ℎ  ( 1 , … ,   ) = (   1  ) (9)

Number of Occurrences of an Integer t as the Powers
Appling (4), we can count the number of occurrences of an integer t as the powers in different terms among all (   1  ) terms of ℎ  ( 1 , … ,   ).
The condition of t is: r ≥ t ≥ 1. Case 1.The terms in which the integer r occurs as the powers on the variables are:  1  , …,    .
Therefore when t = r then the number of occurrences of t is k.
Case 2. When t < r, clearly then r, k ≠ 1.From (4), we get: The number of occurrences of t (10) has some technical terms for some particular values of k, r and t such that the values of these terms are all 0. The particulars in the context are described below.
(i) If m is an integer in (2, …, r) then the product ) is one among r -1 terms of (10).The value of the term is obviously 0 if m > rt + 1.For example, if the triplet (k, r, t) is (12, 7, 4) then the values of the last three terms of (10) where m ∈ (5, 6, 7) are all 0. This implies that if the number of bases in a term of ℎ 7 ( 1 , … ,  12 ) is 5, 6 or 7 then the number of occurrences of 4 as the powers on the bases is 0, or in other words 4 cannot occur as the powers on any of these bases.
(ii) When r > k then the last rk terms have the factors: (   1 ) , …, (   ) in succession such that the values of these rk terms are all 0. In other words, for r > k, the number of occurrences of t is equal to the summation: .

Number of Occurrences of a Variable 𝑥 as the Bases
From Case 1 and Case 2 of Topic 3.2, we get: Total number of bases in all terms of the polynomial The number of occurrences of every variable  ∈ ( 1 , … ,   ) in complete homogeneous symmetric polynomial of degree r in the variables:  1 , … ,   is same.Hence from (11.1), we get: The number of occurrences of a variable  as the bases ) (a) Number of summands in the compositions of n From (5), we get: Number of the summands in the compositions of n for n ≥ 2 = Number of occurrence of t for t = n + Number of occurrences of t for n > t ≥1

=
6) Obviously (6) holds for n = 1 also.(b) A proposition from (5) Proposition 1.If 1 and 2 are the summands in the compositions of 1 and 2 respectively such that (c) Number-number relationship By Pascal's Identity, we get: The above relation implies the following number-number relationship from (4) and (2).Number of occurrences of 1 in C(n, r) = Number of the summands in C(n, r) -Number of the summands in C(n -1, r) we get a number-sum relationship as shown.Number of the summands in the compositions of all n integers: 1, 2, …, n Sum of the summands in the compositions of n. (8) 3. Counting for Complete Homogeneous Symmetric Polynomial:   (  , … ,   )