Symmetry Independent Permutations Without 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 Without Repeatition refer to Arrangements of a Given Number of Distinct Items in a Given Number of Distinct Locations, such that Any One Item can occur Only Once in a given Arrangement. The Count of Symmetry Independent Permutations Without Repeatition is calculated as follows
   1. When Number of Distinct Items is Greater or Equal to the Number of Distinct Locations, the Number of ways the Distinct Locations can be Occupied by Distinct Items can be counted as follows
      
      Let \(N\) be the Number of Distinct Items. Let \(R\) be the Number of Distinct Locations
      
      The \(1^{st}\) Location can be Occupied by \(N\) Items
      
      Once the \(1^{st}\) Location is Occupied, the \(2^{nd}\) Location can be Occupied by \(N-1\) Items
      
      Once the \(1^{st}\) and \(2^{nd}\) Locations are Occupied, the \(3^{rd}\) Location can be Occupied by \(N-2\) Items
      \(\vdots\)
      Once the \(1^{st}, 2^{nd}, 3^{rd}, \cdots, {R-1}^{th}\) Locations are Occupied, the \({R}^{th}\) Location can be Occupied by \(N-R+1\) Items
      
      So by using Product Rule of Fundamental Principle of Counting, the Total Number of ways the \(R\) Distinct Locations can be Occupied by \(N\) Distinct Items can be counted as
      
      \(N \times (N-1) \times (N-2) \times \cdots \times (N-R+1)\)   ...(1)
      
      Multiplying and Dividing the above expression by \((N-R)!\) we get
      
      \(\frac{N \times (N-1) \times (N-2) \times \cdots \times (N-R+1) \times (N-R) \times (N-R-1) \times (N-R-2) \times \cdots \times 2 \times 1} {(N-R) \times (N-R-1) \times (N-R-2) \times \cdots \times 2 \times 1}=\frac{N!}{(N-R)!}=P(N,R)\)   ...(2)
      
      When \(N=R\) (i.e. Number of Distinct Items is Equal to the Number of Distinct Locations) then
      
      \(P(N,N)=\frac{N!}{(N-N)!}=\frac{N!}{0!}=N!\)   ...(3)
   2. When Number of Distinct Locations is Greater or Equal to the Number of Distinct Items, the Number of ways the Distinct Items can be Placed in Distinct Locations can be counted as follows
      
      Let \(N\) be the Number of Distinct Locations. Let \(R\) be the Number of Distinct Items
      
      The \(1^{st}\) Item can be Placed in \(N\) Locations
      
      Once the \(1^{st}\) Item is Placed, the \(2^{nd}\) Item can be Placed in \(N-1\) Locations
      
      Once the \(1^{st}\) and \(2^{nd}\) Items are Placed, the \(3^{rd}\) Item can be Placed in \(N-2\) Locations
      \(\vdots\)
      Once the \(1^{st}, 2^{nd}, 3^{rd}, \cdots, {R-1}^{th}\) Items are Placed, the \({R}^{th}\) Item can be Placed in \(N-R+1\) Locations
      
      So by using Product Rule of Fundamental Principle of Counting, the Total Number of ways the \(R\) Distinct Items can be Placed in \(N\) Distinct Locations can be counted as
      
      \(N \times (N-1) \times (N-2) \times \cdots \times (N-R+1)\)   ...(4)
      
      Multiplying and Dividing the above expression by \((N-R)!\) we get
      
      \(\frac{N \times (N-1) \times (N-2) \times \cdots \times (N-R+1) \times (N-R) \times (N-R-1) \times (N-R-2) \times \cdots \times 2 \times 1} {(N-R) \times (N-R-1) \times (N-R-2) \times \cdots \times 2 \times 1}=\frac{N!}{(N-R)!}=P(N,R)\)   ...(5)
      
      When \(N=R\) (i.e. Number of Distinct Locations is Equal to the Number of Distinct Items) then
      
      \(P(N,N)=\frac{N!}{(N-N)!}=\frac{N!}{0!}=N!\)   ...(6)
3. Equations (3) and (6) give the formula for Counting Symmetry Independent Permutations of Distinct Items Without Repeatition. They state that \(N\) Distinct Items can be Placed in \(N\) Distict Locations in \(N!\) Number of ways. As a corollary, they also state that \(N\) Distinct Items taken All together can be Arranged in a Symmetrically Independent Manner in \(N!\) Number of ways.
   
   Equations (2) and (5) give the formula for Counting Symmetry Independent Permutations of Subset of Distinct Items Without Repeatition. They state that \(N\) Distinct Items can be Placed in \(R\) Distinct Locations (or \(N\) Distinct Locations can be Occupied by \(R\) Distinct Items) (where \(R \leq N\)) in \({\Large\frac{N!}{(N-R)!}}\) Number of ways. As a corollary, they also state that \(N\) Distinct Items when taken \(R\) Items at a time (where \(R \leq N\)) can be Arranged in a Symmetrically Independent Manner in \({\Large\frac{N!}{(N-R)!}}\) Number of ways.
4. An alternate way to define Symmetry Independent Permutations Without Repeatition is that given 2 Sets of Objects \(A\) and \(B\) with Set \(A\) containing \(M\) Number of Distinct Items and Set \(B\) Containing \(N\) Number of Distinct Items, the Symmetry Independent Permutations Without Repeatition give All the Distinct ways \(M\) Items of Set \(A\) can be Simultaneously Paired with \(N\) Items of Set \(B\) . For example, given \(M\) Number of People and \(N\) Number of Hats, Symmetry Independent Permutations Without Repeatition give All the Different ways in which \(M\) Number of People can Simultaneously wear \(N\) Number of Hats. The Count of such Permutations can be calculated as \({\Large\frac{M!}{(M-N)!}}\) (when \(M \geq N\)) or \({\Large\frac{N!}{(N-M)!}}\) (when \(N \geq M\)).