Find the number of positive integers whose digits add up to 42 Find the number of positive integers $$n <9,999,999 $$  for which the sum of the digits in n equals 42.
Can anyone give me any hints on how to solve this?
 A: Let the digits be $d_1,d_2,d_3,d_4,d_5,d_6$, and $d_7$, where we allow leading zeroes so as to make each number in the specified interval a seven-digit integer. You’re looking for all solutions in non-negative integers to 
$$d_1+d_2+d_3+d_4+d_5+d_6+d_7=42\;,$$
with the restriction that $d_k\le 9$ for $k=1,\ldots,7$. Without the restriction this is a standard stars-and-bars problem, whose solution is 
$$\binom{42+7-1}{7-1}=\binom{48}6\;.\tag{1}$$
Both the formula that I used here and a pretty decent explanation/derivation of it can be found at the link. However, $(1)$ includes unwanted solutions in which one or more of the $d_k$ exceeds $9$. To remove these, you can use an inclusion-exclusion argument. This answer shows such an argument in some detail in a smaller problem of this kind.
A: A solution through generating functions:
Imagine a language of words $\newcommand\L{\mathcal{L}}\L=(a^{\{0,9\}}b)^7$, where $a^{\{0,9\}}$ means "any number of $a$'s betwees $0$ and $9$". A word in this language corresponds to $7$ digits in $0,\dots,9$. The sum of the digits is exactly the number of $a$'s in the word. Therefore we want to compute the number of these words such that they contains $42$ occurences of $a$.
Let $C(x)=\sum_{w\in\L} x^{|w|_a}$, where $|w|_a$ is the number of occurences of $a$ in the word $w$. Then the number of words with $42$ letters is exactly the coefficient of $x^{42}$ in $C(x)$, in other words, it is $\frac{1}{42!}\frac{d^{42}}{dx^{42}} C(x)\Bigr|_{x=0}$.
Now, the regular expression for $\L$ is unambigous (every word is obtained exactly once), therefore to get $C(x)$, we simply erase $b$'s and change $a$'s to $x$'s. We get $\bigl(\sum_{i=0}^9x^i\bigr)^7=\Bigl(\frac{x^{10}-1}{x-1}\Bigr)^7$. If you use Maple and put
eval(diff(((x^10-1)/(x-1))^7, x$42), x=0)/factorial(42);

you get the correct answer 209525.
To generalize this, you "only" need to be able to either get a general formula for the derivatives or to be able to express $C(x)$ in a better way.

EDIT: It actually is possible to obtain the closed formula for this. For our series $C(x)$ we have that $C(x)=\Bigl(\frac{1}{1-x}-x^{10}\frac{1}{1-x}\Bigr)^7=\sum_{i=0}^7 \binom{7}{i}(-1)^i x^{10i} \Bigl(\frac{1}{1-x}\Bigr)^7$. Let $b_n$ denote the coefficient of $x^n$ in $\Bigl(\frac{1}{1-x}\Bigr)^7$. We have $b_n=\binom{n+6}{6}$. Then the coefficient of $x^{42}$ in $C(x)$ is $\binom{7}{0}b_{42}-\binom{7}{1}b_{32}+\binom{7}{2}b_{22}-\binom{7}{3}b_{12}+\binom{7}{4}b_2$. Now you can compute on any calculator (or by hand) that this really equals $209525$.

The general formula for $\ell$ digits summing up to $n$ in the base $b$ is:
$$\sum_{i=0}^{\min\{\ell,\lfloor n/b\rfloor\}} (-1)^i \binom{\ell}{i} \binom{n-ib+\ell-1}{\ell-1}.$$
A: Consider the product $(1+x+x^2+x^3+\dots+x^9)^7$. Each choice of a term in each factor corresponds to the choice of a digit in a $7$ digit number. The coefficient of $x^{42}$ corresponds to the number of ways for the sum of those digits to be $42$.
Thus, the answer is the coefficient of $x^{42}$ in
$$
\left(\frac{1-x^{10}}{1-x}\right)^7
=\left(\sum_{j=0}^7\binom{7}{j}(-1)^jx^{10j}\right)\left(\sum_{k=0}^\infty\binom{k+6}{k}x^k\right)
$$
which is given by the Cauchy product of the coefficients of the power series above:
$$
\sum_{j=0}^7(-1)^j\binom{7}{j}\binom{48-10j}{42-10j}=209525
$$
where we take $\binom{n}{k}=0$ when $k\lt0$.
A: I know this is not the answer you are looking for, but I thought it was an amusing one-liner brute force solution (takes less than 1 minute to run) - and may allow you to verify your own result:
python -c "print len([i for i in xrange(10000000) \
                if sum([int(x) for x in str(i)])==42])"

209525
A: Since $0$ has no weight we can consider all numbers as $7$-place decimals. It follows that we have to count the solutions of
$$\sum_{k=1}^7 d_k =42,\qquad 0\leq d_k\leq 9\quad(1\leq k\leq 7)\ .$$
Forgetting about the condition $d_k\leq9$ we have a standard stars-and-bars problem, which has ${42+7-1\choose 7-1}={48\choose6}$ solutions.
Among these solutions there are some with, say, $d_1\geq10$. To count these we count the nonnegative solutions of $\sum_{k=1}^7 d_k=32$ (with the idea to replace in each of these solutions the $d_1$ by $d_1':=d_1+10$ afterwards). 
There are ${38\choose6}$ such solutions.
This number of "forbidden solutions" has to be multiplied by $7$, since each one of the $d_k$ could be $\geq10$. But in this way we count the cases where two $d_k$s are $\geq10$ twice, and it becomes apparent that we run into an inclusion-exclusion scheme. It follows that the total number $N$ we are looking for is given by
$$N={48\choose6}-{7\choose1}{38\choose6}+{7\choose2}{28\choose6}-{7\choose3}{18\choose6}+{7\choose4}{8\choose6}=209525\ .$$
(It is impossible that more than $4$ of the $d_k$ are $\geq10$.)
A: The question is just a variation on this one:
Probability of random integer's digits summing to 12
which provides several neat theoretical solutions. 
For computational evaluation, here is a one-liner in Mathematica (takes about 30 seconds on my Mac):
Count[Map[Total, IntegerDigits[Range[9999998]]], 42]


209525

