Probability of random integer's digits summing to 12

Solution 1:

The problem does not require a spreadsheet. It does not even require paper.

The question is to count the number of integer tuples $\langle a,b,c,d\rangle$ with $a+b+c+d=12$ and $0\le a,b,c,d < 10$. We could enumerate this by choosing $a$ and then counting the tuples $\langle b,c,d\rangle$ with $b+c+d = 12-a$, and recursing, but an easier method is available.

First, note that if we drop the $a,b,c,d < 10$ restriction, the problem is easy. By the stars and bars method, there are $\binom{15}{12} = 455$ tuples that sum to 12.

From these 455 we need to eliminate the ones that contain $10, 11,$ or $12$. Let $t_i$ be the number of tuples where $a =i$ for $i\in\{10,11,12\}$. Clearly, $t_{12} = 1$: the only tuple is $\langle 12, 0,0,0\rangle$. For $a=11$ we need $b+c+d=1$, so exactly one of $b,c,d$ is 1 and the other two are 0, and thus $t_{11} = 3$.

For $a=10$ there are two possibilities. Either $\{b,c,d\} = \{2,0,0\}$ or $\{b,c,d\} = \{1,1,0\}$. In either case there are 3 tuples, so $t_{10} = 6$.

Since at most one of $a,b,c,d$ is greater than 9, the total number of tuples that contain 10, 11, or 12 is $4(t_{10}+t_{11}+t_{12}) = 40$.

Thus the total number of tuples of just 0 through 9, and the answer to the question, is 455 - 40 = 415; the probability is $\frac{415}{9999}$.

Solution 2:

Use generating functions. The generating function for a single digit is:

$$1 + x + \cdots + x^9 = \frac{1-x^{10}}{1-x}.$$

The generating function for the sum of four digits is the fourth power:

$$\frac{(1-x^{10})^4}{(1-x)^4} = (1-x^{10})^4 (1-x)^{-4}.$$

To solve the problem, find the coefficient of $x^{12}$.

$$(1-x^{10})^4 = 1 -4 x^{10} + 6 x^{20} - \cdots$$

So, we only need the coefficients of $x^2$ and $x^{12}$ in $(1-x)^{-4}$ using the generalized binomial theorem. These are $\binom{5}{2} = 10$ and $\binom{15}{12} = 455$. The coefficient of $x^{12}$ is therefore

$$ -4\cdot10 + 1\cdot455 = 415.$$

So the answer is $\frac{415}{9999}$.

Solution 3:

With Mathematica code (FROM 1 TO 9999):

Count[Map[Total, IntegerDigits[Range[9999]]], 12]

415

So: 415/9999

Solution 4:

Just so GAP doesn't get left out. Here's a GAP version to obtain the count:

Number([1..9999],n->Sum(ListOfDigits(n))=12);

which returns 415. So, the probability is $415/9999$.