What is the probability of getting the sum $26$ when $7$ chips are taken out? Suppose you have a bag in which there are $10$ chips numbered $0$ to $9$. You take out a chip at random, note its number and then put it back. This process is done $7$ times and after that the numbers are added. 

What is the probability that the sum is $26$?

The probability of getting any number at any turn is $\frac1{10}$.  Also I know that there are $10^7$ different combinations that can be formed from these numbers. But it would be too inefficient to start counting which ones sum up to $26$. 
I have really no idea what to do. Can this be done in a faster way using combinatorics? If yes, how?
 A: Well, there is an equivalent problem using combinations with repetition. Lets say we want to distribute 26 balls (points) into seven urns (seven draws), such that no urn has more than 10 balls in it. First the number of distribution without restriction is $$ 26+7-1 \choose 6$$
Now for that restriction. We must subtract all possibilities where one urn contains at least ten balls. We can choose this in 7 possible ways. Then we redistribute the remaining balls to get $7 \cdot {16+7-1 \choose 6}$. But in there there is every possibility that two different urns contain at least 10 balls counted twice, therefore we subtract from it the number of possibilities of this event, which is ${7 \choose 2} \cdot {6+7-1 \choose 6}$. Three urns cannot contain ten balls, so we can stop here and get the overall result 
$$  {32 \choose 6 }- 7 \cdot {22 \choose 6}+{7 \choose 2} \cdot {12 \choose 6}=403305$$
A: The number of ways of drawing 7 chits with sum equal to 26 is the coefficient of $x^{26}$ in the expansion
\begin{align*}
(1+x+x^2+\cdots +x^9)^7 &= \left(\frac{1-x^{10}}{1-x}\right)^7 \\
&=(1-7x^{10} + 21x^{20} - \cdots)\left(1+\binom{7}{1}x + \binom{8}{2}x^2 + \cdots\right)
\end{align*}
Coefficient of $x^{26}$ is
$$\binom{32}{26} - 7\cdot \binom{22}{16} + 21\cdot \binom{12}{6} = 403305$$
