Combinations With Unrestricted Repeatition refer to Selection of \(R\) Items from a Set of \(N\) Number of Distinct Type of Items such that Any Item can get repeated any number of times (between 0 to R).
Alternatively Combinations With Unrestricted Repeatition can also be defined in the following 2 ways
Number of ways \(N\) Distinct Type of Items can be Distributed over \(R\) Indistinguishable Items
Number of ways \(N\) Distinct Type of Items can be Combined in Integer Proportions so as to make a Total of R Items
Given the following
Total Number of Distinct Type of Items = \(N\)
Number of Partitions/Separators Required to separate \(N\) Distinct Type of Items= \(N-1\)
Number of Items in a Selection= \(R\)
The Count of Combinations With Unrestricted Repeatition is calculated as the Number of ways the Positions of either the \(N-1\) Separators or the \(R\) Items
can be Selected from the Total Number of Positions given by the Sum of \(N-1\) Separators and R Items (i.e. \((N-1)+R\) or \(N+R-1\)). It is given by the following formula
Count of Combinations With Unrestricted Repeatition = \(C(N+R-1,R)\) = \(C(N+R-1,N-1)\) = \(\frac{(N+R-1)!}{R!(N-1)!}\)
As an example, suppose we are given \(3\) English Alphabets \(A\), \(B\) and \(C\) and we are supposed to Find the Count of All Possible 5 Letter Combinations of these Alphabets.
The following gives \(4\) such possible ways of doing it
where the \(|\) character acts as the Separator/Partition between the type of Alphabets. Please note that any Alphabet can be selected from \(0\) to \(5\) times, making the total Number of Alphabets in any Selection equal to \(5\). In this example
Total Number of Distinct Alphabets \(N\)= \(3\)
Number of Partitions/Separators Required to separate \(N\) Distinct Type of Alphabets= \(N-1\) = \(3-1 = 2\)
Length of each Selection (i.e. Number of Alphabets in each Selection) \(R\) = \(5\)
Sum of Number of Partitions/Separator and Number of Alphabets in each Selection = \(2 +5 = 7\)
Hence, the Count of All Possible 5 Letter Combinations of these Alphabets = \(C(N+R-1,R)\) = \(C(N+R-1,N-1)\) = \(C(3+5-1,5)\) = \(C(7,5)\) = \(C(7,2)\) = \(\frac{7!}{2!5!} = 21\)
Similarly, suppose we are given \(5\) English Alphabets \(A\), \(B\), \(C\), \(D\) and \(E\) and we are supposed to Find the Count of All Possible 3 Letter Combinations of these Alphabets.
The following gives \(4\) such possible ways of doing it
where the \(|\) character acts as the Separator/Partition between the type of Alphabets. Please note that any Alphabet can be selected from \(0\) to \(3\) times, making the total Number of Alphabets in any Selection equal to \(3\). In this example
Total Number of Distinct Alphabets \(N\)= \(5\)
Number of Partitions/Separators Required to separate \(N\) Distinct Type of Alphabets= \(N-1\) = \(5-1 = 4\)
Length of each Selection (i.e. Number of Alphabets in each Selection) \(R\) = \(3\)
Sum of Number of Partitions/Separator and Number of Alphabets in each Selection = \(4 + 3 = 7\)
Hence, the Count of All Possible 3 Letter Combinations of these Alphabets = \(C(N+R-1,R)\) = \(C(N+R-1,N-1)\) = \(C(5+3-1,3)\) = \(C(7,3)\) = \(C(7,4)\) = \(\frac{7!}{3!4!} = 70\)