Groups of numbers that equal a specific sum Find the number of order quadruples of integers $(a(1), a(2), a(3), a(4))$, with $a(n) \geq -4$ such that $a(1)+a(2)+a(3)+a(4)=4$. 
I have no clue where to start. I am familiar with stars and bars, but I don't think I can apply that here.
 A: If you take an 

"ordered quadruple whose entries are each $\ge -4$ summing to $4$" 

and add four to each entry, you get an 

"ordered quadruple of nonnegative integers summing to $20$". 

This correspondence is bijective, so you can instead count the second object, which can be done with stars and bars.
A: First, we note that bars and stars solves the equation for the constraint of $a_n >=0$. Let us examine the common positive constraint, and then extend to the negative constraint.
Consider first the familiar case, where we have $a1 +a2+a3+a4=k $ and take $an>=q$ where $q$ is a positive number. Here, we need to "reserve" $q \times 4$ from k, so each of the $a_n$ has at least $q$. We then subtract $4*q$ from $k$, and solve the new equation $a1 +a2+a3+a4=k-(4\times q)$, without any constraints. We can now apply bars and stars. The intuition behind this is that for every assignment that satisfies the new equation, we can simply add 4 to each of the terms, and end up with a unique assignment of numbers such that $a_n >=q$, and their sum is $k-(4 \times q) +(4\times q)=k$
Turns out, for negative numbers, it is the exact same thing. Let $q=-4$ and solve for $a1 +a2+a3+a4=4+4*4=20$. The intuition here, is that for every solution we obtain such that we have an ordered quadruple of positive integers summing to 20, we can subtract 4 from each of these terms, and now we will have an ordered quadruple summing to 4, where each integer is greater than or equal to -4. This simple translation allows us to "shift" the bars and stars approach to any range we deem appropriate.
