A Sum of All Digit Permutations

Recently, user preshtalwalkar on Twitter posed the following question:

What is the sum of all 5 digit numbers using 1, 2, 3, 4, 5 without repetition?

I will present two solutions: The solution I came up with and a smart one.

My solution: Notice that at each position (the 1's, 10's, 100's, etc.) the sum of the digits is

4! \cdot (1+2+3+4+5) = 24 \cdot 15 = 360,

since there are $4!$ permutations each time a given digit is fixed. Now, summing up the 1's, 10's, 100's, etc. gives

a_5 = 1 \cdot 360 + 10 \cdot 360 + 100 \cdot 360 + \cdots = 11111 \cdot 360 = 3999960

which is the solution.

The smart solution: Note that there are $5!=120$ permutations and at each position the average digit value is 3. Therefore, the answer is

a_5 = 120 \cdot 33333 = 3999960.

This is easily generalized to $n$ -digit numbers, $n \leq 9$ , as

a_n = \frac{n+1}{2} \frac{10^n - 1}{9} n! = \tfrac{1}{18} (10^n - 1) (n + 1)!,

since $(n+1)/2$ is the average digit value and $\sum_{k=0}^{n-1} 10^k=(10^n-1)/9$ , which is a finite sum of a geometric progression. The sequence $a_n$ is A071268 at OEIS.

We can generalize further if we consider $d$ -digit numbers using $1, \ldots, n$ as the digits, $1 \leq d \leq n \leq 9$ . We then have

b_{nd} = \frac{(10^d-1)(n+1)!}{18(n-d)!}

where, of course, $a_n=b_{nn}$ .

janmr blog