How many integer solutions are there for x + y + z + w = 15 if x ≥ 3,y > −2, z ≥ 1, w > −3?

To solve this problem, we can use the method of generating functions or the stars and bars method. However, the stars and bars method is more straightforward in this case. Here are the steps:

1. Adjust the variables to meet the constraints:

   • Since x ≥ 3, let x = x' + 3. Then x' ≥ 0.
   • Since y > -2, let y = y' + 2. Then y' ≥ 0.
   • Since z ≥ 1, let z = z' + 1. Then z' ≥ 0.
   • Since w > -3, let w = w' + 3. Then w' ≥ 0.

2. Substitute these into the original equation:

   • x' + 3 + y' + 2 + z' + 1 + w' + 3 = 15
   • x' + y' + z' + w' = 15 - 3 - 2 - 1 - 3 = 6

3. Now we need to find the number of non-negative integer solutions to x' + y' + z' + w' = 6. This is a standard stars and bars problem. The formula for the number of solutions is (n + k - 1 choose k - 1), where n is the total number (6 in this case) and k is the number of variables (4 in this case).

4. So the number of solutions is (6 + 4 - 1 choose 4 - 1) = (9 choose 3) = 84.

So there are 84 integer solutions to the equation x + y + z + w = 15 with the given constraints.