How many natural numbers are there less than $90000$ that have the sum of digits equal to $8$?

Question

I want to find out how many natural numbers are there less than $90000$ that have the sum of digits equal to $8$.

$x_1+x_2+x_3+x_4+x_5=8$ where $5>x_i>0$.

Relaxing $5>x_i$ for now we get $\binom{7}4$ number of solutions.

Now let $x_1>5$.Clearly there are no solutions for this. So is the answer $\binom{7}4$ ?

@user88595 i made it 900000 but it doesn't make a difference — , Jan 17 '14 at 11:32
Why do you say that $x_i > 0$? Shouldn't it be $x_i \geq 0$? — user121926, Jan 17 '14 at 11:32
So you're requiring that none of the digits should be $0$? In that case, maybe you should consider rewording your question. — user121926, Jan 17 '14 at 11:39
Your question is unclear the set of natural numbers less than 90000 is 1,2,3,...10,11,...30...89999. lots of these numbers have zeros in them. So if you don't want zeros you will have to tell your rules for creating a number. I'm assuming here that $x_1$ is the first digit of your number for example. — Warren Hill, Jan 17 '14 at 12:06
Why do you say there is no solution with $x_1>5$? For example, 62000 seems to be a perfectly good solution, since $6+2+0+0+0=8$. — hmakholm left over Monica, Jan 17 '14 at 13:19

score 1 · Answer 1 · answered Jan 17 '14 at 13:19

I also got 35. Did it the old fashioned way so there may be an error. Note we can't pick any digit to be 5 either because there would be 4 more digits to pick each of which must be at least 1 so the total sum becomes > 8.

1) Pick the first digit to be 4 then all other digits must be 1

4 1 1 1 1

there are 5 positions where 4 can go so there are 5 of these numbers.

2) Pick the first digit to be 3 then we cannot pick another 3 because we would go over. Pick the second digit 2

3 2

now , all remaining digits must be 1

3 2 1 1 1

keeping 3 fixed , there are four positions for 2. Since there are five positions for 3 we use the multiplication principle to get 5×4 = 20 of these numbers.

3) Pick the first digit to be 2. There is no need to pick 3 as any digit because that has been done already above at 2). Pick the second digit to be 2

2 2

the sum is already 4. now there are three digits remaining to pick and one of them must be 2

2 2 2

we are forced to pick 1 for the remaining digits

2 2 2 1 1

let's work with the 1's switching their position. We can use the same argument as before to get 5×4 = 20 but now we divide by two because the digits switching position are equal. There are 10 of these numbers.

There is no need to pick the first digit to be 1 because all possible configurations have been counted already. We are done counting.

5 + 20 + 10 = 35

score 1 · Answer 2 · answered Jan 17 '14 at 18:31

One way to do this is to use what's called the stars and bars theorem. This states that the number of integer solutions to $$ y_1+y_2+\cdots+y_n = m, \quad y_i\ge0 $$ is $$\binom{m+n-1}{m} $$ For example, with $n=2, m=3$ we could view this as the number of ways of arranging three stars and one (i.e., $n-1$) bar separating the stars, so we'd have correspondences like these $$\begin{align} y_1=1,\ y_2=2&\ \longleftrightarrow\ *|**\\ y_1=3,\ y_2=0&\ \longleftrightarrow\ ***| \end{align}$$ Now this doesn't quite solve your problem, since you require that the solutions be strictly greater than zero, but if we take your problem $$ x_1+x_2+\cdots+x_5 = 8, \quad x_i\color{red}{>}0 $$ and let $y_i=x_i-1$, we'd then have the equivalent $$ x_1+x_2+\cdots+x_5 = 8\Longrightarrow (y_1+1)+(y_2+1)+\cdots+(y_5+1)=8 $$ giving us the form we need: $$ y_1+y_2+\cdots+y_5 = 3, \quad y_i\color{red}{\ge} 0 $$ which the theorem says has $\binom{3+5-1}{3}=\binom{7}{3}=35$ solutions, as expected.

How many natural numbers are there less than $90000$ that have the sum of digits equal to $8$?

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