Distribution problem where |a|, |b|, |c|, and |d| are at most 10. Check my work? How many ways can a+b+c+d=18, where a,b,c,d are integers such that $|a|,\ |b|,\ |c|,\ |d|$
 are each at most 10?
This is what I have so far. If all four numbers have the restriction -10 =< a, b, c, d, =< 10, than we can add 11 to each number, where
w = a+11,
x = b+11,
y = c+11,
z = d+11
w+x+y+z = 29. Would it be $C(28,3)$ or did I overcount some numbers? 
 A: This is a balls and urns problem solvable using a combination of inclusion-exclusion and stars&bars
$|a|\leq 10 \Rightarrow -10\leq a \leq 10$.  Make a change of variable $x_1 = a+10\Rightarrow 0\leq x_1 \leq 20$
Do similarly to the other letters, so we are at $x_1 + x_2 + x_3 + x_4 = (a+10) + (b+10) + (c+10) + (d+10) = (a+b+c+d)+40 = 18+40 = 58$
Number of ways in which no upper bound conditions are violated = number of ways without upper bound condition - number of ways at least one condition is violated + number of ways two conditions are violated - number of ways 3 conditions are violated + number of ways 4 conditions are violated.
To violate one upper bound condition (say for example $x_1$) then $x_1>20\Rightarrow x_1\geq 21$ make a change of variable $y_1 = x_1 - 21$.  Then $0\leq y_1$ and $y_1+x_2+x_3+x_4 = 37$.  As there were 4 choices for which was the offending $x$, we multiply by 4.
Do similarly for the case of two violating terms.  Note for 3 or 4 violating terms, it is impossible.
$ = \binom{58+4-1}{4-1} - 4\cdot \binom{37+4-1}{4-1} + 6\cdot\binom{16+4-1}{4-1} - 0 + 0 = 2284$

It should be mentioned in case you haven't seen stars&bars, to find the number of integer solutions of $x_1 + x_2 + \dots + x_r = n$ where each $x_i\geq 0$, there are $\binom{n+r-1}{r-1}$ number of solutions.
