counting: sum of 1..1000015, how many numbers are there?

**dayscott** · December 10th 2009, 06:11 PM

i have an advanced counting problem:

given the intergers 1..1000015, how many intergers are there, whose sum of digits equals 15?

i solved this problem brute force in ocaml, but this wasn't really the task ^^ (solution is 13992)

i think that i have to use the formula for the problem which can be illustrated by asking: " how many possibilites are there to give n coins to m children?"
formula:
${m+n-1 \choose n-1}$

i have no clue how to apply the formula to my current problem.

**qmech** · December 10th 2009, 07:39 PM

I can see 2 ways to count the # of solutions. In both of these, I count from 0 to 999,999 and manually count the number of ways from 1,000,000 to 1,000,015.

First, list the number of ways to write 15 = 9+6 =8+7, = 5+5+5, etc. Then find the number of ways to put these digits into the 6 slots that form the numbers from 0-999,999. for instance, there are 6 choose 2 ways to put a 9 and a 6 into a list of 6 slots. The other slots are 0. This way seems a bit tedious.

The 2nd way is to consider the number 15 as 15 coins, and the division between digits as lines that you insert between coins. There should be 5 lines, making 20 items total. Then the number of ways to insert 5 lines into 20 slots is 20 choose 5. The resulting number is the number of coins between lines. For instance, line, coin, coin, line, line, coin is the number 0201. The bad point about this method is that you might have 15 coins between 2 lines, indicating a digit with value 15. There is of course no such digit base 10.

**awkward** · December 12th 2009, 04:49 PM

Originally Posted by dayscott

i have an advanced counting problem:

given the intergers 1..1000015, how many intergers are there, whose sum of digits equals 15?

i solved this problem brute force in ocaml, but this wasn't really the task ^^ (solution is 13992)

i think that i have to use the formula for the problem which can be illustrated by asking: " how many possibilites are there to give n coins to m children?"
formula:
${m+n-1 \choose n-1}$

i have no clue how to apply the formula to my current problem.

Here is an approach via generating functions.

By inspection, there are no solutions in the range 1,000,000 to 1,000,015.
So consider the equation

$x_1 + x_2 + x_3 + \dots + x_6 = r$

where $0 \leq x_i \leq 9$ for all i, with $x_i$ an integer.

We want the number of solutions when r = 15. But let's see if we can solve the more general problem. So let the number of solutions be $a_r$ , and define $f(x) = \sum_{r=0}^{\infty} a_r x^r$ . It's easy to see that
$f(x) = (1 + x + x^2 + \dots + x^9)^6 = [(1-x^{10}) (1-x)^{-1}]^6 = (1-x^{10})^6 \cdot (1-x)^{-6}$ .
(If you have trouble seeing this, think about what happens when you expand $(1 + x + x^2 + \dots + x^9)^6)$ .

Applying the binomial theorem,
$f(x) = \sum_i (-1)^i \binom{6}{i} x^{10i} \cdot \sum_j \binom{6+j-1}{j} x^j$ .
Extracting the coefficient of $x^{15}$ , we find that

$a_{15} = \binom{20}{15} - \binom{6}{1} \binom{10}{5} = 13,992$

**Plato** · December 12th 2009, 05:33 PM

Originally Posted by awkward

By inspection, there are no solutions in the range 1,000,000 to 1,000,015.

ok

**Defunkt** · December 12th 2009, 06:58 PM

.

**dayscott** · December 13th 2009, 10:36 AM

excercise has been solved:

you just count ${20 \choose 15} = 15504$ than you subtract ${9 \choose 5}+ {8 \choose 4}+{7 \choose 3}+{6 \choose 2}+{5 \choose 1} + {4 \choose 0}$

explanation of ${9 \choose 5} which is {5-1+5 \choose 5}$ is: you have got 5 which you have to distribute to 5 digits.

thx for help!

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