How many natural numbers are there less than $90000$ that have the sum of digits equal to $8$? 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 $\binom{7}4$ ?
 A: 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
A: 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. 
