Sum of all numbers that can be formed with permutations of n digits - GeeksforGeeks

Sum of all numbers that can be formed with permutations of n digits - GeeksforGeeks
Given n distinct digits (from 0 to 9), find sum of all n digit numbers that can be formed using these digits. It is assumed that numbers formed with leading 0 are allowed.

Total numbers that can be formed using n digits is total number of permutations of n digits, i.e factorial(n). Now, since the number formed is a n-digit number, each digit will appear factorial(n)/n times at each position from least significant digit to most significant digit. Therefore, sum of digits at a position = (sum of all the digits) * (factorial(n)/n).

Considering the example digits as 1 2 3

factorial(3)/3 = 2

Sum of digits at least significant digit = (1 + 2 + 3) * 2 = 12

Similarly sum of digits at tens, hundreds place is 12. 
(This sum will contribute as 12 * 100)

Similarly sum of digits at tens, thousands place is 12. 
(This sum will contribute as 12 * 1000)

Required sum of all numbers = 12 + (10 * 12) + (100 * 12) = 1332

int getSum(int arr[],int n)

{

    // calculate factorial

    int fact = factorial(n);

 

    // sum of all the given digits at different

    // positions is same and is going to be stored

    // in digitsum.

    int digitsum = 0;

    for (int i=0; i<n; i++)

        digitsum += arr[i];

    digitsum *= (fact/n);

 

    // Compute result (sum of all the numbers)

    int res = 0;

    for (int i=1, k=1; i<=n; i++)

    {

        res  += (k*digitsum);

        k = k*10;

    }

 

    return res;

}

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