Consider a digital combination lock which has x dirty or worn buttons; that is, we know we have some combination containing x different characters, but we do not know how many times each character is used. We want to know how many possible combinations there are if the code is n digits long (if we don't know what n is, to crack the code we would try all combinations starting with the smallest possible n, then progress to combinations with one added digit until the code is found). This is given by the expression below.
The first quantity in the equation is the total possible combinations of an n digit long code, with x possible characters available. The total number of combinations includes, for example, codes where only one character is used n times. Since we know that exactly x characters are used, unless x is one, the number of combinations is not that great. When x is one, only this first term is needed, and the number of combinations is always one.
The second quantity is needed for x > 1, and eliminates all combinations where only one character is used. The third quantity is needed for x > 2, and it enumerates all combinations n long with only 2 characters used. Similarly, the third expression is needed for x > 3, and it enumerates combinations with 3 characters. Similar expressions are added for x > 4, 5, 6..., and it can be seen that these expressions follow a simple pattern.
The summation and first quotient in the third quantity enumerates the number of permutations of two different characters in an n long sequence, with j of one character, and (n - j) of the other character. The summation is to adjust j so that every possible quantity of each character is used. The quotient multiplying this quantity takes into account the possible combinations of two characters available to be used in the enumeration.
The fourth and later quantities operate similar to the third. For the case x = n, the expression reduces to a factorial, where N = x! = n! can be used instead.
Below is a table enumerating the combinations with sequences as long as 10 digits using up to 10 different characters.
| Length of combination (n) | |||||||||||
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ||
| Number of characters used (x) | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 0 | 2 | 6 | 14 | 30 | 62 | 126 | 254 | 510 | 1022 | |
| 3 | 0 | 0 | 6 | 36 | 150 | 540 | 1,806 | 5,796 | 18,150 | 55,980 | |
| 4 | 0 | 0 | 0 | 24 | 240 | 1,560 | 8,400 | 40824 | 186,480 | 818,520 | |
| 5 | 0 | 0 | 0 | 0 | 120 | 1,800 | 16,800 | 126,000 | 834,120 | 5,103,000 | |
| 6 | 0 | 0 | 0 | 0 | 0 | 720 | 15,120 | 191,520 | 1,905,120 | 16,435,440 | |
| 7 | 0 | 0 | 0 | 0 | 0 | 0 | 5,040 | 141,120 | 2,328,480 | 29,635,200 | |
| 8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 40,320 | 1,451,520 | 30,240,000 | |
| 9 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 362,880 | 16,329,600 | |
| 10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 3,628,800 | |
It can be seen that if one were to pick a combination n digits long for their lock, it would be wise to pick a combination that uses x characters such that x maximizes the number of combinations available if n is publicly known. If a certain lock is known to use codes 5 digits long, for example, it would be best to pick a combination using 4 characters. This way, when the buttons become worn or dirty so that the characters used are known, it leaves the maximum possible number of combinations that one would have to guess before cracking the code.