Hex color combinatorics | cryptic crossroads

Last updated Sep 12, 2025

The question

Given a sum n, how many hexadecimal color codes are there with digits that add to n?

I wondered this while writing about strategies for The Password Game. You win by creating a password that, among other silly things, (a) includes the hex code of a randomly chosen color and (b) has digits that add to 25, no more and no less. Since you can get a new random color as many times as you want, I wanted to estimate how long it would take to find one that kept the sum of digits below 25. I wrote a script to simulate the problem and got a good estimate that way, but I noticed there was a pattern to how likely it was for a hex code to have any given sum, and became curious about what that pattern was.

Background & research

Since combinatorics problems often use gambling as an example, I tried thinking of my question in terms of a game where I roll six die, representing the six characters in a hex code. Indeed, a formula has been found for the number of ways to get n points on a dice roll, where k is the number of dice and s is the number of sides per die:

\sum_{j = 0}^{⌊ (n - k) / s ⌋} (- 1)^{j} (\binom{k}{j}) (\binom{n - s j - 1}{k - 1})

I later learned that generically, this is a formula for compositions: the number of sequences of k integers 1 through s that add to n.

"Wait, what do those symbols mean?"

$(\binom{n}{k})$ is a combination.
$\sum_{n = 0}^{k} n$ is a summation.
$⌊ n ⌋$ is the floor function.

The answer

The number of hex codes with a sum n can be calculated as follows:

\sum_{k = 0}^{6} (\binom{6}{k}) 7^{6 - k} \sum_{j = 0}^{⌊ (n - k) / 9 ⌋} (- 1)^{j} (\binom{k}{j}) (\binom{n - 9 j - 1}{k - 1})

Let's break that down

$\sum_{k = 0}^{6}$ : Any number of characters k in a hex code, from none of them to all six, can be a digit 1 through 9.

$(\binom{6}{k})$ : Those digits can appear in any k of the six positions.

$7^{6 - k}$ : There are seven choices; 0, A, B, C, D, E, and F; for each of the $6 - k$ remaining characters.

We know the rest gives us the number of sequences of k non-zero digits that add up to n. To understand why, we can use the stars and bars method.