# Count of ways to rearrange N digits and M alphabets keeping all alphabets together

Last Updated : 01 Mar, 2023

Given two positive integers N and M representing the count of distinct digits and alphabets respectively in a string, the task to count the number of ways to rearrange the characters of the string such that all the alphabets are adjacent.

Examples:

Input: N = 2, M = 2
Output: 12
Explanation: Possible ways to rearrange characters of a string such that all alphabets are adjacent: { {N1N2M2M1, N2N1M2M1, N2N1M1M2, N1N2M1M2, M2M1N1N2, M1M2N2N1, M1M2N1N2, M2M1N2N1, N1M1M2N2, N2M1M2N1, N1M2M1N2, N2M2M1N1} }.

Input: N = 2, M = 4
Output: 144

Naive Approach: The simplest approach to solve this problem to make a string consisting of N distinct numeric characters and M distinct alphabets. Now, generate all possible permutations of the string and check if all the alphabets of the string are adjacent or not. If found to be true, then increment the count. Finally, print the count obtained.

Time Complexity: O((N + M)!)
Auxiliary Space: O(N + M)

Efficient Approach: The problem can be solved based on the following observations:

Since all alphabets are adjacent, therefore consider all the alphabets as a single character.
Therefore, the total count of ways to rearrange the string by considering all alphabets to a single character = ((N + 1)!) * (M!)

Follow the below steps to solve the problem:

1. Calculate the factorial of N + 1 say, X and factorial of M say, Y.
2. Finally, print the value of (X * Y).

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include` `using` `namespace` `std;`   `// Function to find the factorial` `// of the given number` `int` `fact(``int` `n)` `{` `    ``int` `ans = 1;`   `    ``for``(``int` `i = 2; i <= n; i++)` `        ``ans = ans * i;` `    ``return` `ans;` `}`   `// Function to count ways to rearrange` `// characters of the string such that ` `// all alphabets are adjacent.` `int` `findComb(``int` `N, ``int` `M)` `{`   `    ``// Stores factorial of (N + 1)` `    ``int` `x = fact(N + 1);`   `    ``// Stores factorial of ` `    ``int` `y = fact(M);` `    ``return` `(x * y);`   `}`   `// Driver Code` `int` `main()` `{` `  `  `// Given a and b` `int` `N = 2;` `int` `M = 2;``// Function call` `cout<

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Function to find the factorial` `// of the given number` `static` `int` `fact(``int` `n)` `{` `    ``int` `ans = ``1``;`   `    ``for``(``int` `i = ``2``; i <= n; i++)` `        ``ans = ans * i;` `        `  `    ``return` `ans;` `}`   `// Function to count ways to rearrange` `// characters of the String such that ` `// all alphabets are adjacent.` `static` `int` `findComb(``int` `N, ``int` `M)` `{` `    `  `    ``// Stores factorial of (N + 1)` `    ``int` `x = fact(N + ``1``);`   `    ``// Stores factorial of ` `    ``int` `y = fact(M);` `    ``return` `(x * y);` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `  `  `    ``// Given a and b` `    ``int` `N = ``2``;` `    ``int` `M = ``2``;` `    `  `    ``// Function call` `    ``System.out.print(findComb(N, M));` `}` `}`   `// This code is contributed by umadevi9616`

## Python3

 `# Python program of the above approach`   `import` `math`   `# Function to find the factorial` `# of the given number`     `def` `fact(a):` `    ``return` `math.factorial(a)`   `# Function to count ways to rearrange` `# characters of the string such that ` `# all alphabets are adjacent.` `def` `findComb(N, M):`   `    ``# Stores factorial of (N + 1)` `    ``x ``=` `fact(N ``+` `1``)` `    `  `    ``# Stores factorial of ` `    ``y ``=` `fact(M)` `    ``return` `(x ``*` `y)`     `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``# Given a and b` `    ``N ``=` `2` `    ``M ``=` `2`   `    ``# Function call` `    ``print``(findComb(N, M))`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{` `    ``// Function to find the factorial` `// of the given number` `static` `int` `fact(``int` `n)` `{` `    ``int` `ans = 1;`   `    ``for``(``int` `i = 2; i <= n; i++)` `        ``ans = ans * i;` `    ``return` `ans;` `}`   `// Function to count ways to rearrange` `// characters of the string such that ` `// all alphabets are adjacent.` `static` `int` `findComb(``int` `N, ``int` `M)` `{`   `    ``// Stores factorial of (N + 1)` `    ``int` `x = fact(N + 1);`   `    ``// Stores factorial of ` `    ``int` `y = fact(M);` `    ``return` `(x * y);`   `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `  `  `// Given a and b` `int` `N = 2;` `int` `M = 2;``// Function call` `Console.Write(findComb(N, M));`   `}` `}`   `// This code is contributed by bgangwar59.`

## Javascript

 ``

Output:

`12`

Time Complexity: O(N + M)
Auxiliary Space: O(1)

### Approach 3 :

The approach used in this code is based on combinatorics and the factorial of N and M.

The function fact(n) calculates the factorial of a given number, which is the product of all integers from 1 to n. This function takes O(n) time as it is a simple for loop that iterates from 2 to n, multiplying the result by the current number in each iteration.

The function findComb(N, M) calculates the total number of ways to rearrange the characters such that all alphabets are adjacent, by multiplying the factorial of N+1 and M. The factorial of N+1 is used to account for the number of ways to arrange the digits around the alphabets, and the factorial of M is used to account for the number of ways to arrange the alphabets together.

## C++

 `#include` `using` `namespace` `std;`   `// Function to find the factorial` `// of the given number` `int` `fact(``int` `n)` `{` `    ``int` `ans = 1;`   `    ``for``(``int` `i = 2; i <= n; i++)` `        ``ans = ans * i;` `    ``return` `ans;` `}`   `// Function to count ways to rearrange` `// characters of the string such that` `// all alphabets are adjacent.` `int` `findComb(``int` `N, ``int` `M)` `{`   `    ``// Stores factorial of (N + 1)` `    ``int` `x = fact(N + 1);`   `    ``// Stores factorial of` `    ``int` `y = fact(M);` `    ``return` `(x * y);`   `}`   `// Driver Code` `int` `main()` `{` `    ``// Given a and b` `    ``int` `N = 2;` `    ``int` `M = 2;` `    ``// Function call` `    ``cout<

## Java

 `public` `class` `Main {`   `    ``// Function to find the factorial` `    ``// of the given number` `    ``static` `int` `fact(``int` `n) {` `        ``int` `ans = ``1``;` `        ``for` `(``int` `i = ``2``; i <= n; i++) {` `            ``ans = ans * i;` `        ``}` `        ``return` `ans;` `    ``}`   `    ``// Function to count ways to rearrange` `    ``// characters of the string such that` `    ``// all alphabets are adjacent.` `    ``static` `int` `findComb(``int` `N, ``int` `M) {` `        ``// Stores factorial of (N + 1)` `        ``int` `x = fact(N + ``1``);`   `        ``// Stores factorial of` `        ``int` `y = fact(M);` `        ``return` `(x * y);` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``// Given a and b` `        ``int` `N = ``2``;` `        ``int` `M = ``2``;` `        ``// Function call` `        ``System.out.println(findComb(N, M));` `    ``}` `}`

## Python3

 `# Python code implemetation for the above approach`   `# Function to find the factorial of the given number` `def` `fact(n):` `    ``ans ``=` `1` `    ``for` `i ``in` `range``(``2``, n``+``1``):` `        ``ans ``=` `ans ``*` `i` `    ``return` `ans`   `# Function to count ways to rearrange characters of ` ` ``# the string such that all alphabets are adjacent.` `def` `findComb(N, M):` `    ``# Stores factorial of (N + 1)` `    ``x ``=` `fact(N ``+` `1``)`   `    ``# Stores factorial of` `    ``y ``=` `fact(M)` `    ``return` `(x ``*` `y)`     `# Given N and M` `N ``=` `2` `M ``=` `2`   `# Function call` `print``(findComb(N, M))`   `# This code is contributed by karthik.`

## C#

 `// C# code addition for the above approach` `using` `System;`   `public` `class` `GFG {`   `  ``// Function to find the factorial of the given number` `  ``static` `int` `fact(``int` `n)` `  ``{` `    ``int` `ans = 1;` `    ``for` `(``int` `i = 2; i <= n; i++) {` `      ``ans = ans * i;` `    ``}` `    ``return` `ans;` `  ``}`   `  ``// Function to count ways to rearrange characters of the` `  ``// string such that all alphabets are adjacent.` `  ``static` `int` `findComb(``int` `N, ``int` `M)` `  ``{` `    ``// Stores factorial of (N + 1)` `    ``int` `x = fact(N + 1);`   `    ``// Stores factorial of` `    ``int` `y = fact(M);` `    ``return` `(x * y);` `  ``}`   `  ``static` `public` `void` `Main()` `  ``{`   `    ``// Code` `    ``// Given a and b` `    ``int` `N = 2;` `    ``int` `M = 2;` `    ``// Function call` `    ``Console.WriteLine(findComb(N, M));` `  ``}` `}`   `// This code is contributed by sankar.`

## Javascript

 `// Function to find the factorial` `// of the given number` `function` `fact(n) {` `  ``let ans = 1;` `  ``for` `(let i = 2; i <= n; i++) {` `    ``ans = ans * i;` `  ``}` `  ``return` `ans;` `}`   `// Function to count ways to rearrange` `// characters of the string such that` `// all alphabets are adjacent.` `function` `findComb(N, M) {` `  ``// Stores factorial of (N + 1)` `  ``let x = fact(N + 1);`   `  ``// Stores factorial of` `  ``let y = fact(M);` `  ``return` `(x * y);` `}`   `// Given a and b` `let N = 2;` `let M = 2;` `// Function call` `console.log(findComb(N, M));` `// this code is contributed by writer`

Output

`12`

Time complexity : O(n) as it only involves one for loop in the fact(n) function and the rest of the calculations are constant time operations. This approach is simple and efficient as it only requires a single for loop to calculate the factorial of N and M and the rest of the calculations are simple arithmetic operations.

Auxiliary Space : O(1)

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