# Count of distinct permutations of length N having no similar adjacent characters

Given an integer N, the task is to calculate the total number of distinct permutations of length **N**, consisting only of letters ‘a’, ‘b’, and ‘c’, with repetitions allowed, such that no two adjacent characters are the same.

Input:N = 3Output:12Explanation:

Possible permutations satisfying the required conditions are {aba, abc, aca, acb, bac, bab, bca, bcb, cac, cab, cba, cbc}Input:N = 5Output:48

**Approach:**

Following observations need to be made to solve the given problem:

- Let us fix the first letter as
**‘a’**. - Now, the 2
^{nd}letter can either be**‘b’**or**‘c’**, which leaves 2 ways to fill the second letter. - Similarly, the third letter can also be filled in two ways. If the character in 2
^{nd}position is ‘b’, the third character can either be ‘a’ or ‘c’. If the character in 2^{nd}position is ‘c’, the third character can either be ‘a’ or ‘b’. - Similarly, for all the remaining positions, there will always be two possibilities depending on the character in the previous position. Therefore, the total number of possible permutation if ‘a’ occupies the first position is
**1*2*2*2…*2 = 1 * 2**.^{N – 1} - Therefore, the total number of permutations, considering the first character can also be ‘b’ or ‘c’ as well, is
**3 * 2**.^{N – 1}

Below is the implementation of the above approach:

## C++

`// C++ Program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to print the number of` `// permutations possible` `int` `countofPermutations(` `int` `N)` `{` ` ` `return` `int` `(3 * ` `pow` `(2, N - 1));` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 5;` ` ` `cout << countofPermutations(N);` ` ` `return` `0;` `}` |

## Java

`// Java program to implement` `// the above approach` `class` `GFG{` `// Function to print the number of` `// permutations possible` `static` `int` `countofPermutations(` `int` `N)` `{` ` ` `return` `(` `int` `)(` `3` `* Math.pow(` `2` `, N - ` `1` `));` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `5` `;` ` ` `System.out.print(countofPermutations(N));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python3 program to implement` `# the above approach` `# Function to print the number of` `# permutations possible` `def` `countofPermutations(N):` ` ` ` ` `return` `int` `((` `3` `*` `pow` `(` `2` `, N ` `-` `1` `)));` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `N ` `=` `5` `;` ` ` `print` `(countofPermutations(N));` `# This code is contributed by amal kumar choubey` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{` `// Function to print the number of` `// permutations possible` `static` `int` `countofPermutations(` `int` `N)` `{` ` ` `return` `(` `int` `)(3 * Math.Pow(2, N - 1));` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 5;` ` ` ` ` `Console.Write(countofPermutations(N));` `}` `}` `// This code is contributed by Amit Katiyar` |

## Javascript

`<script>` `// Javascript Program to implement` `// the above approach` `// Function to print the number of` `// permutations possible` `function` `countofPermutations(N)` `{` ` ` `return` `parseInt(3 * Math.pow(2, N - 1));` `}` `// Driver Code` `var` `N = 5;` `document.write( countofPermutations(N));` `</script>` |

**Output:**

48

**Time Complexity: **O(logN) **Auxiliary Space: **O(1)

Attention reader! Don’t stop learning now. Participate in the **Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students**.