Using Generating Functions to Find Combinations, Count of Combinations and Count of Linear Permutations

1. Given Any Set of Distinct Items, the Generating Functions can be used to Find and Count Any and All of ways the Items can be Selected in Any Combination and Find the Count of All Possible Linear Permutations for those Combinations.
2. The following example demonstrates how Generating Functions are constructed for Finding and Counting Combinations and Counting Linear Permutations.
   
   Suppose we have to make from \(0\) to \(3\) Letter Combinations of English Alphabet \(A\), the same can be done in 4 ways
   
   1 using No Letters = \(A^0\)
   
   1 using Single Letter \(A\) = \(A^1 = A\)
   
   1 using 2 Letters of \(A\) = \(A^2 = AA\)
   
   1 using 3 Letters of \(A\) = \(A^3 = AAA\)
   
   \(\therefore\) As per Sum Rule of Fundamental Principle of Counting Total Number of Combinations having \(0\) to \(3\) Letters of \(A\) = \(A^0 + A^1 + A^2 + A^3=4\)   ...(1)
   
   Similarly, Suppose we have to make from \(0\) to \(4\) Letter Combinations of English Alphabet \(B\), the same can be done in 5 ways
   
   1 using No Letters = \(B^0\)
   
   1 using Single Letter \(B\) = \(B^1 = B\)
   
   1 using 2 Letters of \(B\) = \(B^2 = BB\)
   
   1 using 3 Letters of \(B\) = \(B^3 = BBB\)
   
   1 using 4 Letters of \(B\) = \(B^4 = BBBB\)
   
   \(\therefore\) As per Sum Rule of Fundamental Principle of Counting Total Number of Combinations having \(0\) to \(4\) Letters of \(B\) = \(B^0 + B^1 + B^2 + B^3 + B^4 = 5\)   ...(2)
   
   Similarly, Suppose we have to make from \(0\) to \(2\) Letter Combinations of English Alphabet \(C\), the same can be done in 3 ways
   
   1 using No Letters = \(C^0\)
   
   1 using Single Letter \(C\) = \(C^1 = C\)
   
   1 using 2 Letters of \(C\) = \(C^2 = CC\)
   
   \(\therefore\) As per Sum Rule of Fundamental Principle of Counting Total Number of Combinations having \(0\) to \(2\) Letters of \(C\) = \(C^0 + C^1 + C^2 = 3\)   ...(3)
   
   So, as per Product Rule of Fundamental Principle of Counting, the Total Count of Combinations having \(0\) to \(3\) Letters of \(A\), \(0\) to \(4\) Letters of \(B\) and \(0\) to \(2\) Letters of \(C\) can be calculated as follows
   
   Total Count of Combinations = Combinations having \(0\) to \(3\) Letters of \(A\) \(\times\) Combinations having \(0\) to \(4\) Letters of \(B\) \(\times\) Combinations having \(0\) to \(2\) Letters of \(C\) = \(4 \times 5 \times 2 = 60\)
   
   The Product Rule of Fundamental Principle of Counting can also be used to find the actual Combinations by multiplying equations (1), (2) and (3) as follows
   
   \((A^0 + A^1 + A^2 + A^3)\times (B^0 + B^1 + B^2 + B^3 + B^4)\times (C^0 + C^1 + C^2)\)
   
   \(=(1 + A + A^2 + A^3)\times (1 + B + B^2 + B^3 + B^4)\times (1 + C + C^2)\)
   
   \(=(1 + A + A^2 + A^3 + B + B^2 + B^3 + B^4 + AB + AB^2 + AB^3 + \\ AB^4 + A^2B + A^2B^2 + A^2B^3 + A^2B^4 + A^3B + A^3B^2 + \\ A^3B^3 + A^3B^4 )\times (1 + C + C^2)\)
   
   \(=1 + A + A^2 + A^3 + B + B^2 + B^3 + B^4 + C + C^2 + \\ AB + AB^2 + AB^3 + AB^4 + A^2B + A^2B^2 + A^2B^3 + A^2B^4 + \\ A^3B + A^3B^2 + A^3B^3 + A^3B^4 + AC + A^2C + A^3C + BC + \\ B^2C + B^3C + B^4C + ABC + AB^2C + AB^3C + AB^4C + \\ A^2BC + A^2B^2C + A^2B^3C + A^2B^4C + A^3BC + A^3B^2C + \\ A^3B^3C + A^3B^4C + AC^2 + A^2C^2 + A^3C^2 + BC^2 + B^2C^2 + \\ B^3C^2 + B^4C^2 + ABC^2 + AB^2C^2 + AB^3C^2 + AB^4C^2 + A^2BC^2 + \\ A^2B^2C^2 + A^2B^3C^2 + A^2B^4C^2 + A^3BC^2 + A^3B^2C^2 + A^3B^3C^2 + A^3B^4C^2\)   ...(4)
   
   The expression (4) above gives the Generating Function for finding Any Combination of Any Length and Composition having \(0\) to \(3\) Letters of \(A\), \(0\) to \(4\) Letters of \(B\) and \(0\) to \(2\) Letters of \(C\). Each term of this Generating Function gives a Unique Combination of Alphabets \(A\), \(B\) and \(C\).
   
   Generating Function to Find Count of Linear Permutations of Combinations of Alphabets \(A\), \(B\) and \(C\) given in expression (4) can by found by Multiplying each Term / Combination by the Number of Linear Permutations that the Term / Combination can have as given below
   
   \(=1 + (\frac{1!}{1!})A + (\frac{2!}{2!})A^2 + (\frac{3!}{3!})A^3 + (\frac{1!}{1!})B + (\frac{2!}{2!})B^2 + (\frac{3!}{3!})B^3 + (\frac{4!}{4!})B^4 + (\frac{1!}{1!})C + (\frac{2!}{2!})C^2 + \\ (\frac{2!}{1!1!})AB + (\frac{3!}{1!2!})AB^2 + (\frac{4!}{1!3!})AB^3 + (\frac{5!}{1!4!})AB^4 + (\frac{3!}{2!1!})A^2B + (\frac{4!}{2!2!})A^2B^2 + (\frac{5!}{2!3!})A^2B^3 + (\frac{6!}{2!4!})A^2B^4 + \\ (\frac{4!}{3!1!})A^3B + (\frac{5!}{3!2!})A^3B^2 + (\frac{6!}{3!3!})A^3B^3 + (\frac{7!}{3!4!})A^3B^4 + (\frac{2!}{1!1!})AC + (\frac{3!}{2!1!})A^2C + (\frac{4!}{3!1!})A^3C + (\frac{2!}{1!1!})BC + \\ (\frac{3!}{2!1!})B^2C + (\frac{4!}{3!1!})B^3C + (\frac{5!}{4!1!})B^4C + (\frac{3!}{1!1!1!})ABC + (\frac{4!}{1!2!1!})AB^2C + (\frac{5!}{1!3!1!})AB^3C + (\frac{6!}{1!4!1!})AB^4C + \\ (\frac{4!}{2!1!1!})A^2BC + (\frac{5!}{2!2!1!})A^2B^2C + (\frac{6!}{2!3!1!})A^2B^3C + (\frac{7!}{2!4!1!})A^2B^4C + (\frac{5!}{3!1!1!})A^3BC + (\frac{6!}{3!2!1!})A^3B^2C + \\ (\frac{7!}{3!3!1!})A^3B^3C + (\frac{8!}{3!4!1!})A^3B^4C + (\frac{3!}{1!2!})AC^2 + (\frac{4!}{2!2!})A^2C^2 + (\frac{5!}{3!2!})A^3C^2 + (\frac{3!}{1!2!})BC^2 + (\frac{4!}{2!2!})B^2C^2 + \\ (\frac{5!}{3!2!})B^3C^2 + (\frac{6!}{4!2!})B^4C^2 + (\frac{4!}{1!1!2!})ABC^2 + (\frac{5!}{1!2!2!})AB^2C^2 + (\frac{6!}{1!3!2!})AB^3C^2 + (\frac{7!}{1!4!2!})AB^4C^2 + (\frac{5!}{2!1!2!})A^2BC^2 + \\ (\frac{6!}{2!2!2!})A^2B^2C^2 + (\frac{7!}{2!3!2!})A^2B^3C^2 + (\frac{8!}{2!4!2!})A^2B^4C^2 + (\frac{6!}{3!1!2!})A^3BC^2 + (\frac{7!}{3!2!2!})A^3B^2C^2 + (\frac{8!}{3!3!2!})A^3B^3C^2 + (\frac{9!}{3!4!2!})A^3B^4C^2\)   ...(5)
   
   
3. To determine the Number of Combinations of a Particular Length, the Generating Function given in expresssion (4) can be simplified by replacing the three Alphabets \(A\), \(B\) and \(C\) with a Single Alphabet (for e.g. \(X\)) as follows
   
   \(1 + A + A^2 + A^3 + B + B^2 + B^3 + B^4 + C + C^2 + \\ AB + AB^2 + AB^3 + AB^4 + A^2B + A^2B^2 + A^2B^3 + A^2B^4 + \\ A^3B + A^3B^2 + A^3B^3 + A^3B^4 + AC + A^2C + A^3C + BC + \\ B^2C + B^3C + B^4C + ABC + AB^2C + AB^3C + AB^4C + \\ A^2BC + A^2B^2C + A^2B^3C + A^2B^4C + A^3BC + A^3B^2C + \\ A^3B^3C + A^3B^4C + AC^2 + A^2C^2 + A^3C^2 + BC^2 + B^2C^2 + \\ B^3C^2 + B^4C^2 + ABC^2 + AB^2C^2 + AB^3C^2 + AB^4C^2 + A^2BC^2 + \\ A^2B^2C^2 + A^2B^3C^2 + A^2B^4C^2 + A^3BC^2 + A^3B^2C^2 + A^3B^3C^2 + A^3B^4C^2\)
   
   \(=1 + X + X^2 + X^3 + X + X^2 + X^3 + X^4 + X + X^2 + \\ XX + XX^2 + XX^3 + XX^4 + X^2X + X^2X^2 + X^2X^3 + X^2X^4 + \\ X^3X + X^3X^2 + X^3X^3 + X^3X^4 + XX + X^2X + X^3X + XX + \\ X^2X + X^3X + X^4X + XXX + XX^2X + XX^3X + XX^4X + \\ X^2XX + X^2X^2X + X^2X^3X + X^2X^4X + X^3XX + X^3X^2X + \\ X^3X^3X + X^3X^4X + XX^2 + X^2X^2 + X^3X^2 + XX^2 + X^2X^2 + \\ X^3X^2 + X^4X^2 + XXX^2 + XX^2X^2 + XX^3X^2 + XX^4X^2 + X^2XX^2 + \\ X^2X^2X^2 + X^2X^3X^2 + X^2X^4X^2 + X^3XX^2 + X^3X^2X^2 + X^3X^3X^2 + X^3X^4X^2\)   ...(6)
   
   \(=1 + X + X^2 + X^3 + X + X^2 + X^3 + X^4 + X + X^2 + \\ X^2 + X^3 + X^4 + X^5 + X^3 + X^4 + X^5 + X^6 + \\ X^4 + X^5 + X^6 + X^7 + X^2 + X^3 + X^4 + X^2 + \\ X^3 + X^4 + X^5 + X^3 + X^4 + X^5 + X^6 + \\ X^4 + X^5 + X^6 + X^7 + X^5 + X^6 + \\ X^7 + X^8 + X^3 + X^4 + X^5 + X^3 + X^4 + \\ X^5 + X^6 + X^4 + X^5 + X^6 + X^7 + X^5 + \\ X^6 + X^7 + X^8 + X^6 + X^7 + X^8 + X^9\)   ...(7)
   
   \(=1 + 3X + 6X^2 + 9X^3 + 11X^4 + 11X^5 + 9X^6 + 6X^7 + 3X^8 + X^9\)   ...(8)
   
   The expression (8) above gives the concise form of the Generating Function given in expression (4). Each term of this Polynomial represents the Combination of a Particular Length given by the Power of the Alphabet \(X\). The Co-efficients of each term gives the Total Count of Combinations of Length specified by the Power of the Alphabet \(X\).
4. To determine the Number of Permutations of a Particular Length, the Generating Function given in expression (5) can be simplified by replacing the three Alphabets \(A\), \(B\) and \(C\) with a Single Alphabet (for e.g. \(X\)) as follows
   
   \(1 + (\frac{1!}{1!})A + (\frac{2!}{2!})A^2 + (\frac{3!}{3!})A^3 + (\frac{1!}{1!})B + (\frac{2!}{2!})B^2 + (\frac{3!}{3!})B^3 + (\frac{4!}{4!})B^4 + (\frac{1!}{1!})C + (\frac{2!}{2!})C^2 + \\ (\frac{2!}{1!1!})AB + (\frac{3!}{1!2!})AB^2 + (\frac{4!}{1!3!})AB^3 + (\frac{5!}{1!4!})AB^4 + (\frac{3!}{2!1!})A^2B + (\frac{4!}{2!2!})A^2B^2 + (\frac{5!}{2!3!})A^2B^3 + (\frac{6!}{2!4!})A^2B^4 + \\ (\frac{4!}{3!1!})A^3B + (\frac{5!}{3!2!})A^3B^2 + (\frac{6!}{3!3!})A^3B^3 + (\frac{7!}{3!4!})A^3B^4 + (\frac{2!}{1!1!})AC + (\frac{3!}{2!1!})A^2C + (\frac{4!}{3!1!})A^3C + (\frac{2!}{1!1!})BC + \\ (\frac{3!}{2!1!})B^2C + (\frac{4!}{3!1!})B^3C + (\frac{5!}{4!1!})B^4C + (\frac{3!}{1!1!1!})ABC + (\frac{4!}{1!2!1!})AB^2C + (\frac{5!}{1!3!1!})AB^3C + (\frac{6!}{1!4!1!})AB^4C + \\ (\frac{4!}{2!1!1!})A^2BC + (\frac{5!}{2!2!1!})A^2B^2C + (\frac{6!}{2!3!1!})A^2B^3C + (\frac{7!}{2!4!1!})A^2B^4C + (\frac{5!}{3!1!1!})A^3BC + (\frac{6!}{3!2!1!})A^3B^2C + \\ (\frac{7!}{3!3!1!})A^3B^3C + (\frac{8!}{3!4!1!})A^3B^4C + (\frac{3!}{1!2!})AC^2 + (\frac{4!}{2!2!})A^2C^2 + (\frac{5!}{3!2!})A^3C^2 + (\frac{3!}{1!2!})BC^2 + (\frac{4!}{2!2!})B^2C^2 + \\ (\frac{5!}{3!2!})B^3C^2 + (\frac{6!}{4!2!})B^4C^2 + (\frac{4!}{1!1!2!})ABC^2 + (\frac{5!}{1!2!2!})AB^2C^2 + (\frac{6!}{1!3!2!})AB^3C^2 + (\frac{7!}{1!4!2!})AB^4C^2 + (\frac{5!}{2!1!2!})A^2BC^2 + \\ (\frac{6!}{2!2!2!})A^2B^2C^2 + (\frac{7!}{2!3!2!})A^2B^3C^2 + (\frac{8!}{2!4!2!})A^2B^4C^2 + (\frac{6!}{3!1!2!})A^3BC^2 + (\frac{7!}{3!2!2!})A^3B^2C^2 + (\frac{8!}{3!3!2!})A^3B^3C^2 + (\frac{9!}{3!4!2!})A^3B^4C^2\)
   
   \(=1 + (\frac{1!}{1!})X + (\frac{2!}{2!})X^2 + (\frac{3!}{3!})X^3 + (\frac{1!}{1!})X + (\frac{2!}{2!})X^2 + (\frac{3!}{3!})X^3 + (\frac{4!}{4!})X^4 + (\frac{1!}{1!})X + (\frac{2!}{2!})X^2 + \\ (\frac{2!}{1!1!})XX + (\frac{3!}{1!2!})XX^2 + (\frac{4!}{1!3!})XX^3 + (\frac{5!}{1!4!})XX^4 + (\frac{3!}{2!1!})X^2X + (\frac{4!}{2!2!})X^2X^2 + (\frac{5!}{2!3!})X^2X^3 + (\frac{6!}{2!4!})X^2X^4 + \\ (\frac{4!}{3!1!})X^3X + (\frac{5!}{3!2!})X^3X^2 + (\frac{6!}{3!3!})X^3X^3 + (\frac{7!}{3!4!})X^3X^4 + (\frac{2!}{1!1!})XX + (\frac{3!}{2!1!})X^2X + (\frac{4!}{3!1!})X^3X + (\frac{2!}{1!1!})XX + \\ (\frac{3!}{2!1!})X^2X + (\frac{4!}{3!1!})X^3X + (\frac{5!}{4!1!})X^4X + (\frac{3!}{1!1!1!})XXX + (\frac{4!}{1!2!1!})XX^2X + (\frac{5!}{1!3!1!})XX^3X + (\frac{6!}{1!4!1!})XX^4X + \\ (\frac{4!}{2!1!1!})X^2XX + (\frac{5!}{2!2!1!})X^2X^2X + (\frac{6!}{2!3!1!})X^2X^3X + (\frac{7!}{2!4!1!})X^2X^4X + (\frac{5!}{3!1!1!})X^3XX + (\frac{6!}{3!2!1!})X^3X^2X + \\ (\frac{7!}{3!3!1!})X^3X^3X + (\frac{8!}{3!4!1!})X^3X^4X + (\frac{3!}{1!2!})XX^2 + (\frac{4!}{2!2!})X^2X^2 + (\frac{5!}{3!2!})X^3X^2 + (\frac{3!}{1!2!})XX^2 + (\frac{4!}{2!2!})X^2X^2 + \\ (\frac{5!}{3!2!})X^3X^2 + (\frac{6!}{4!2!})X^4X^2 + (\frac{4!}{1!1!2!})XXX^2 + (\frac{5!}{1!2!2!})XX^2X^2 + (\frac{6!}{1!3!2!})XX^3X^2 + (\frac{7!}{1!4!2!})XX^4X^2 + (\frac{5!}{2!1!2!})X^2XX^2 + \\ (\frac{6!}{2!2!2!})X^2X^2X^2 + (\frac{7!}{2!3!2!})X^2X^3X^2 + (\frac{8!}{2!4!2!})X^2X^4X^2 + (\frac{6!}{3!1!2!})X^3XX^2 + (\frac{7!}{3!2!2!})X^3X^2X^2 + (\frac{8!}{3!3!2!})X^3X^3X^2 + (\frac{9!}{3!4!2!})X^3X^4X^2\)   ...(9)
   
   \(=1 + (\frac{1!}{1!})X + (\frac{2!}{2!})X^2 + (\frac{3!}{3!})X^3 + (\frac{1!}{1!})X + (\frac{2!}{2!})X^2 + (\frac{3!}{3!})X^3 + (\frac{4!}{4!})X^4 + (\frac{1!}{1!})X + (\frac{2!}{2!})X^2 + \\ (\frac{2!}{1!1!})X^2 + (\frac{3!}{1!2!})X^3 + (\frac{4!}{1!3!})X^4 + (\frac{5!}{1!4!})X^5 + (\frac{3!}{2!1!})X^3 + (\frac{4!}{2!2!})X^4 + (\frac{5!}{2!3!})X^5 + (\frac{6!}{2!4!})X^6 + \\ (\frac{4!}{3!1!})X^4 + (\frac{5!}{3!2!})X^5 + (\frac{6!}{3!3!})X^6 + (\frac{7!}{3!4!})X^7 + (\frac{2!}{1!1!})X^2 + (\frac{3!}{2!1!})X^3 + (\frac{4!}{3!1!})X^4 + (\frac{2!}{1!1!})X^2 + \\ (\frac{3!}{2!1!})X^3 + (\frac{4!}{3!1!})X^4 + (\frac{5!}{4!1!})X^5 + (\frac{3!}{1!1!1!})X^3 + (\frac{4!}{1!2!1!})X^4 + (\frac{5!}{1!3!1!})X^5 + (\frac{6!}{1!4!1!})X^6 + \\ (\frac{4!}{2!1!1!})X^4 + (\frac{5!}{2!2!1!})X^5 + (\frac{6!}{2!3!1!})X^6 + (\frac{7!}{2!4!1!})X^7 + (\frac{5!}{3!1!1!})X^5 + (\frac{6!}{3!2!1!})X^6 + \\ (\frac{7!}{3!3!1!})X^7 + (\frac{8!}{3!4!1!})X^8 + (\frac{3!}{1!2!})X^3 + (\frac{4!}{2!2!})X^4 + (\frac{5!}{3!2!})X^5 + (\frac{3!}{1!2!})X^3 + (\frac{4!}{2!2!})X^4 + \\ (\frac{5!}{3!2!})X^5 + (\frac{6!}{4!2!})X^6 + (\frac{4!}{1!1!2!})X^4 + (\frac{5!}{1!2!2!})X^5 + (\frac{6!}{1!3!2!})X^6 + (\frac{7!}{1!4!2!})X^7 + (\frac{5!}{2!1!2!})X^5 + \\ (\frac{6!}{2!2!2!})X^6 + (\frac{7!}{2!3!2!})X^7 + (\frac{8!}{2!4!2!})X^8 + (\frac{6!}{3!1!2!})X^6 + (\frac{7!}{3!2!2!})X^7 + (\frac{8!}{3!3!2!})X^8 + (\frac{9!}{3!4!2!})X^9\)   ...(10)
   
   \(=1 + (\frac{1!}{1!} + \frac{1!}{1!} + \frac{1!}{1!})X + ( \frac{2!}{2!} + \frac{2!}{2!} + \frac{2!}{2!} + \frac{2!}{1!1!} + \frac{2!}{1!1!} + \frac{2!}{1!1!}) X^2 + \\ (\frac{3!}{3!} + \frac{3!}{3!} + \frac{3!}{1!2!} + \frac{3!}{2!1!} + \frac{3!}{2!1!} + \frac{3!}{2!1!} + \frac{3!}{1!1!1!} + \frac{3!}{1!2!} + \frac{3!}{1!2!}) X^3 + \\ (\frac{4!}{4!} + \frac{4!}{1!3!} + \frac{4!}{2!2!} + \frac{4!}{3!1!} + \frac{4!}{3!1!} + \frac{4!}{3!1!} + \frac{4!}{1!2!1!} + \frac{4!}{2!1!1!} + \frac{4!}{2!2!} + \frac{4!}{2!2!} + \frac{4!}{1!1!2!}) X^4 + \\ (\frac{5!}{1!4!} + \frac{5!}{2!3!} + \frac{5!}{3!2!} + \frac{5!}{4!1!} + \frac{5!}{1!3!1!} + \frac{5!}{2!2!1!} + \frac{5!}{3!1!1!} + \frac{5!}{3!2!} + \frac{5!}{3!2!} + \frac{5!}{1!2!2!} + \frac{5!}{2!1!2!}) X^5 + \\ (\frac{6!}{2!4!} + \frac{6!}{3!3!} + \frac{6!}{1!4!1!} + \frac{6!}{2!3!1!} + \frac{6!}{3!2!1!} + \frac{6!}{4!2!} + \frac{6!}{1!3!2!} + \frac{6!}{2!2!2!} + \frac{6!}{3!1!2!} ) X^6 + \\ (\frac{7!}{3!4!} + \frac{7!}{2!4!1!} + \frac{7!}{3!3!1!} + \frac{7!}{1!4!2!} + \frac{7!}{2!3!2!} + \frac{7!}{3!2!2!})X^7 + (\frac{8!}{3!4!1!} + \frac{8!}{2!4!2!} + \frac{8!}{3!3!2!}) X^8 + (\frac{9!}{3!4!2!}) X^9\)   ...(11)
   
   The expression (11) above gives the concise form of the Generating Function given in expression (5). Each term of this Polynomial represents the Permutation of a Particular Length given by the Power of the Alphabet \(X\). The Co-efficients of each term gives the Total Count of Permutations of Length specified by the Power of the Alphabet \(X\).
5. Based on the example given above, the following gives the Steps to for Calculating the Generating Functions for Combinations and Linear Permutations
   1. Identify the Distinct Objects from which the Selections / Combinations / Linear Permutations are to be made and assign a different Variable to each such Object.
   2. Based on the Minimum and Maximum times each Object can be present in the Selections / Combinations, construct the Polynomial for each Variable as demonstrated in equations (1), (2) and (3) above.
   3. Generating Function for All the Possible Combinations of the Variables can be found by Multipling the Polynomials of all the Variables (as demonstrated through expression (4)). This Generating Function can be used for finding Any Combination of Any Length and Composition of Objects.
   4. Generating Function to Find Count of Linear Permutations of Combinations of the Objects / Variables can by found by Multiplying each Term / Combination of the Generating Function of Combinations by the Number of Linear Permutations that the Term / Combination can have (as demonstrated through expression (5)).
   5. To determine the Number of Combinations of a Particular Length, the Generating Function of Combinations can be simplified by replacing the all the Variables with a Single Variable as demonstrated through expressions (6), (7) and (8). Each term of this Simplified Generating Function represents the Combination of a Particular Length given by the Power of the Variable and the Co-efficients of each term gives the Total Count of Combinations of Length specified by the Power of the Variable.
   6. To determine the Number of Permutations of a Particular Length, the Generating Function of Linear Permutations can be simplified by replacing the all the Variables with a Single Variable as demonstrated through expressions (9), (10) and (11). Each term of this Simplified Generating Function represents the Permutation of a Particular Length given by the Power of the Variable and the Co-efficients of each term gives the Total Count of Permutations of Length specified by the Power of the Variable.