Final answer:
The number of solutions to the equation x1 + x2 + x3 + x4 + x5 + x6 = 25 is determined by different restrictions placed on the values of xi. Using the concept of stars and bars, we can find the number of non-negative integer solutions for each restriction. The number of solutions without any restrictions is 142,506, with the restriction xi ≥ 3 for all i being 2002, with 3 ≤ x1 ≤ 10 being 84,024, and with 3 ≤ x1 ≤ 10 and 2 ≤ x2 ≤ 7 being 8,568.
Step-by-step explanation:
In this problem, we are looking for the number of non-negative integer solutions to the equation x1 + x2 + x3 + x4 + x5 + x6 = 25, with different restrictions on the values of xi.
1) There are no other restrictions:
In this case, we can use the concept of stars and bars to find the number of solutions. The formula for the number of non-negative integer solutions to this equation is (n+k-1)C(k-1), where n is the target sum (25) and k is the number of variables (6).
So, the number of solutions is (25+6-1)C(6-1) = 30C5 = 142,506.
2) xi ≥ 3 for i = 1, 2, 3, 4, 5, 6:
In this case, we can subtract 3 from each xi to make them non-negative. Then, we have x1' + x2' + x3' + x4' + x5' + x6' = 9, where x1', x2', x3', x4', x5', x6' ≥ 0.
Using the stars and bars formula again, the number of solutions is (9+6-1)C(6-1) = 14C5 = 2002.
3) 3 ≤ x1 ≤ 10:
In this case, we can subtract 3 from x1 to make it non-negative. Then, we have x1' + x2 + x3 + x4 + x5 + x6 = 22, where x1' = x1 - 3 and x1', x2, x3, x4, x5, x6 ≥ 0.
Using the stars and bars formula, the number of solutions is (22+6-1)C(6-1) = 27C5 = 84,024.
4) 3 ≤ x1 ≤ 10 and 2 ≤ x2 ≤ 7:
In this case, we can subtract 3 from x1 and 2 from x2 to make them non-negative. Then, we have x1' + x2' + x3 + x4 + x5 + x6 = 13, where x1' = x1 - 3 and x2' = x2 - 2, and x1', x2', x3, x4, x5, x6 ≥ 0.
Using the stars and bars formula, the number of solutions is (13+6-1)C(6-1) = 18C5 = 8568.