Symmetry Independent Permutations With Repeatition

1. Symmetry Independent Permutations refer to Arrangements of a Given Number of Objects/Items Without taking into consideration the Symmetry of the Arrangements.
2. Symmetry Independent Permutations With Repeatition refer to All Distinct Arrangements of \(N\) Number of Distinct Items in a sequence of a given Length \(R\), such that Any Item can occur any number of times (between \(0\) and \(R\)) in any given Arrangement.
3. The Count of Total Number of Symmetry Independent Permutations With Repeatition can be calculated as follows
   
   The \(1^{st}\) Position can be occupied by any of \(N\) Possible Items
   
   The \(2^{nd}\) Position can also be occupied by any of \(N\) Possible Items
   
   The \(3^{rd}\) Position can also be occupied by any of \(N\) Possible Items
   \(\vdots\)
   The \(R^{th}\) Position can also be occupied by any of \(N\) Possible Items
   
   Therefore as per Product Rule of Fundamental Principle of Counting, the Count of Total Number of Possible Arangements are
   
   \(N \times N \times N \times\) (Repeated \(R\) times)= \(N^R\)
4. The following are 2 examples of Symmetry Independent Permutations With Repeatition
   1. The total number of 3 Letter Words that can be formed by using 26 characters of English Alphabets are \({26}^3\).
   2. The total number of 5 Digit Numbers that can be formed by using the 10 digits \(0\) to \(9\) are \({10}^5\).
5. The Symmetry Independent Permutations With Repeatition of \(N\) Number of Distinct Items Taking \(R\) Number of Items at a time can be expressed in form of Multinomial Expansion. For example, Symmetry Independent Permutations With Repeatition of 3 Items \(a\), \(b\) and \(c\) taking 4 Items at a time can be given as Multinomial Expansion of \((a+b+c)^4\) as follows
   
   \((a+b+c)^4 = {\Large\frac{4!}{4!}} a^4 + {\Large\frac{4!}{3!1!}}a^3b + {\Large\frac{4!}{3!1!}}a^3c + {\Large\frac{4!}{2!2!}}a^2b^2 + {\Large\frac{4!}{2!1!1!}}a^2bc + {\Large\frac{4!}{2!2!}}a^2c^2 + {\Large\frac{4!}{1!3!}}ab^3 + {\Large\frac{4!}{1!2!1!}}ab^2c + {\Large\frac{4!}{1!1!2!}}abc^2 + {\Large\frac{4!}{1!3!}}ac^3 + {\Large\frac{4!}{4!}}b^4 + {\Large\frac{4!}{3!1!}}b^3c + {\Large\frac{4!}{2!2!}}b^2c^2 + {\Large\frac{4!}{1!3!}}bc^3 + {\Large\frac{4!}{4!}}c^4\)   ...(7)
   
   \(\Rightarrow (a+b+c)^4 = a^4 + 4a^3b + 4a^3c + 6a^2b^2 + 12a^2bc + 6a^2c^2 + 4ab^3 + 12 ab^2c + 12abc^2 + 4ac^3 + b^4 + 4b^3c + 6b^2c^2 + 4bc^3 + c^4\)   ...(8)
   
   The Variables of any particular term of the Multinomial Expansion (as given by equations (7) and (8)) specify a Particular Combination in which 4 Items can be chosen and the Corresponding Coefficients specify the Number of Symmetry Independent Permutations Possible for that Combination of 4 Items. The Total Count of Symmetry Independent Permutations With Repeatition is \(3^4=81\), which is same as the Sum of the Coefficients of all the terms of the Multinomial Expansion as given by equation (8) above.
   
   The Coefficient of Each Term in equations (7) and (8) give the Count of Symmetry Independent Permutations of \(N\) Number of Items of which \(P\) Items are of Type \(a\), \(Q\) Items are of Type \(b\) and \(R\) Items are of Type \(c\) and is calculated as \({\Large\frac{N!}{P!Q!R!}}\). The following gives its derivation
   
   Number of ways \(N\) Items can be arranged = \(N!\)
   
   Number of ways \(P\) Number of Item \(a\) can be arranged = \(P!\)
   
   Number of ways \(Q\) Number of Item \(b\) can be arranged = \(Q!\)
   
   Number of ways \(R\) Number of Item \(c\) can be arranged = \(R!\)
   
   Therefore as per Product Rule of Fundamental Principle of Counting, the Total Number of Arrangements Possible of Items \(A\), \(B\) and \(C\) taken together are = \(P!\times Q!\times R! = P!Q!R!\)
   
   Since for any given arrangement of \(N\) Items there are \(P!Q!R!\) Number of ways \(P\) Number of Item \(a\), \(Q\) Number of Item \(b\) and \(R\) Number of Item \(c\) can be arranged, therefore Number of Permutations of \(N\) Items that get merged into a Single Permutation is also \(P!Q!R!\) (as they are indistinguishable from each other).
   
   Hence, Net Total Number of Permutations With Repeatition of Items \(a\), \(b\) and \(c\)
   
   = \(\frac{Total\hspace{.1cm}Number\hspace{.1cm}of\hspace{.1cm}Permutations\hspace{.1cm}of\hspace{.1cm}N\hspace{.1cm}Items} {Total\hspace{.1cm}Number\hspace{.1cm}of\hspace{.1cm}Permutations\hspace{.1cm}Getting\hspace{.1cm}Merged\hspace{.1cm}into\hspace{.1cm}Single\hspace{.1cm}Permutation\hspace{.1cm}Because\hspace{.1cm}of\hspace{.1cm}Repeatition\hspace{.1cm}of\hspace{.1cm}Item\hspace{.1cm}A,\hspace{.1cm}B\hspace{.1cm}and\hspace{.1cm}C}={\Large\frac{N!}{P!Q!R!}}\)
   
   In equation (7) above, value of \(N=4\) and values of \(P\), \(Q\) and \(R\) are powers of \(a\), \(b\) and \(c\) respectively in the different terms of the Multinomial Expansion.
   
   As another example, Symmetry Independent Permutations With Repeatition of 2 Items \(a\) and \(b\) taking 5 Items at a time can be given as Multinomial Expansion of \((a+b)^5\) as follows
   
   \((a+b)^5 = {\Large\frac{5!}{5!}} a^5 + {\Large\frac{5!}{4!1!}} a^4b + {\Large\frac{5!}{3!2!}} a^3b^2 + {\Large\frac{5!}{2!3!}} a^2b^3 + {\Large\frac{5!}{1!4!}} ab^4 + {\Large\frac{5!}{5!}} b^5\)   ...(9)
   
   \(\Rightarrow (a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5\)   ...(10)
   
   The Variables of any particular term of the Multinomial Expansion (as given by equations (9) and (10)) specify a Particular Combination in which 5 Items can be chosen and the Corresponding Coefficients specify the Number of Symmetry Independent Permutations Possible for that Combination of 5 Items. The Total Count of Symmetry Independent Permutations With Repeatition is \(2^5=32\), which is same as the Sum of the Coefficients of all the terms of the Multinomial Expansion as given by equation (10) above.
   
   The Coefficient of Each Term in equations (9) and (10) give the Count of Symmetry Independent Permutations of \(N\) Number of Items of which \(P\) Items are of Type \(a\) and \(Q\) Items are of Type \(b\) and is calculated as \({\Large\frac{N!}{P!Q!}}\). In equation (9) above, value of \(N=5\) and values of \(P\) and \(Q\) are powers of \(a\) and \(b\) respectively in the different terms of the Multinomial Expansion.