# Probability of random integer's digits summing to 12

What is the probability that a random integer between 1 and 9999 will have digits that sum to 12?

As a user suggested, I could make a spreadsheet and count them, but is there a quicker way to do this?

• For a range like this, why not make a spreadsheet and count them? – Ross Millikan Apr 25 '13 at 16:52
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• I vote against closing this question. – MJD Apr 25 '13 at 17:40
• I also vote against closing this question. – Zev Chonoles Apr 25 '13 at 22:02
• @TheChaz I am very disappointed to see valid questions like this being closed for strange reasons, e.g. "too localized". Ditto for analogous recent votes. – Math Gems Apr 26 '13 at 15:50

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 $\frac{415}{9999}$.

• Note that with the range starting at 1, leading zeros are explicitly OK. This makes it easier. – Ross Millikan Apr 25 '13 at 18:12
• There are several features of this problem that make it easier than it seems at first: leading zeroes are acceptable, so $a,b,c,$ and $d$ are symmetric. The required sum is close to 9, so only a few integer tuples need to be subtracted, and no inclusion-exclusion argument is needed. Recognizing such features when they appear can be an important aspect of solving this type of problem. – MJD Apr 25 '13 at 18:27
• Does not even require paper?? :) Neither did mine – wolfies Apr 25 '13 at 19:17
• The problem does not require an integer either. What is the probability that randomly chosen combination of four numbers in the range 0 to 9 adds up to to twelve. – Kaz Apr 25 '13 at 21:53
• @Kaz Probability is 0, because if you permit $(a,b,c,d) \in \mathbb{R}^4$, even restricting the possible range to between 0 and 9, $a+b+c+d=12$ is still a three-dimentional figure. That's like asking what the probability that a random point in a plane happens to lie on a given line, or the probability that a randomly chosen number just happens to be exactly the same as a given number. Without restricting the digits to integers, the problem becomes trivial. – AJMansfield Apr 26 '13 at 1:24

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 $\frac{415}{9999}$.

• where do 5,2,15 and 12 come from in ncr(5, 2) and ncr(15, 12)? – michaelsnowden Sep 8 '15 at 17:46
• @michaelsnowden The coefficient of $x^2$ in $(1 - x)^{-4}$ is $(-1)^2\binom{-4}{2} = \binom{4 + 2 - 1}{2}$, because of the binomial theorem which says that the coefficient of $z^n$ in $(1+z)^n$ is $\binom{n}{r}$, and the fact that $\binom{-n}{r} = (-1)^r\binom{n + r - 1}{r}$. Similarly, the coefficient of $x^{12}$ in $(1-x)^{-4}$ is $(-1)^{12}\binom{-4}{12} = \binom{4 + 12 - 1}{12}$. – ShreevatsaR Jan 5 '19 at 21:16
• In general, if we're looking for $k$ random digits summing to $n$, by this answer we're looking for the coefficient of $x^n$ in $(1-x^{10})^k (1-x)^{-k}$. This is going to be $$\sum_{10r+s=n} (-1)^{r+s} \binom{k}{r} \binom{-k}{s} = \sum_{10r+s=n} (-1)^{r} \binom{k}{r} \binom{k+s-1}{s}$$ if I'm not mistaken. – ShreevatsaR Jan 5 '19 at 21:24

With Mathematica code (FROM 1 TO 9999):

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


415

So: 415/9999

• As an alternative approach: Probability[x == 12, x \[Distributed] Total[IntegerDigits[Range[9999]], {2}]] directly yields 415/9999. – Sasha Apr 25 '13 at 19:05
• Wow, "FrankenLisp". – Kaz Apr 25 '13 at 21:54

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 $415/9999$.

• Go GAP! Good to see it's still around and alive. – Peter K. Apr 25 '13 at 20:57

With PARI/GP code

Q(n)=if(n<10,n,n%10 + Q(n\10))

sum(i=1,9999,Q(i)==12)/9999

I obtain $$\frac{415}{9999}.$$