Final answer:
To find the number of integer solutions for the equation x1 + x2 + x3 + x4 = 28, we can use the concept of stars and bars. There are 35960 solutions for 0 ≤ xi ≤ 12, 270725 solutions for -10 ≤ xi ≤ 20, 10696 solutions for 0 ≤ xi ≤ 6, x1 ≤ 1 and x3 ≤ 15, and 435 solutions for x3 ≤ 15 and x4 ≤ 21.
Step-by-step explanation:
To find the number of integer solutions for the equation x1 + x2 + x3 + x4 = 28, we can use the concept of stars and bars.
a) For 0 ≤ xi ≤ 12, we can treat each xi as a star and the 28 as bars. So we have 28 bars and 4 stars. The number of ways to arrange these 28 bars and 4 stars is equal to choosing 4 positions out of the 32 (28 bars + 4 stars). This can be calculated using the binomial coefficient:
C(32, 4) = (32!)/(4!*(32-4)!) = 35960.
b) For -10 ≤ xi ≤ 20, we need to shift the range of xi by adding 10 to each xi. This gives us x1' + x2' + x3' + x4' = 48 (x1' = x1 + 10, x2' = x2 + 10, x3' = x3 + 10, x4' = x4 + 10). Now we can use the same concept and calculate the number of solutions:
C(52, 4) = (52!)/(4!*(52-4)!) = 270725.
c) For 0 ≤ xi ≤ 6, x1 ≤ 1, and x3 ≤ 15, we have three ranges to consider. First, we need to find the number of solutions for x1 + x2' + x3' + x4' = 28 (where x2' = x2, x3' = x3, x4' = x4). This can be calculated as C(31, 3) = (31!)/(3!*(31-3)!) = 4495. Then, we need to find the number of solutions for x1 = 0 and x1 = 1. For x1 = 0, we have x2' + x3' + x4' = 28. This can be calculated as C(28, 3) = 3276. For x1 = 1, we have x2' + x3' + x4' = 27. This can be calculated as C(27, 3) = 2925. Adding the solutions together, we get 4495 + 3276 + 2925 = 10696.
d) For x3 ≤ 15 and x4 ≤ 21, we can treat x3 and x4 as a single variable (let's call it x'). Now we have x1 + x2 + x' = 28. This can be calculated as C(30, 2) = (30!)/(2!*(30-2)!) = 435.