Final answer:
There are different restrictions for the number of solutions to the equation x₁ + x₂ + x₃ + x₄ + x₅ + x₆ = 25. 1) There are no other restrictions: The number of solutions is 142,506. 2) x₁ ≥ 3 for i = 1, 2, 3, 4, 5, 6: There are no solutions. 3) 3 ≤ x₁ ≤ 10: The total number of solutions is (C(22, 5) + C(21, 5) + ... + C(15, 5)). 4) 3 ≤ x₁ ≤ 10 and 2 ≤ x₂ ≤ 7: The total number of solutions is (C(15, 5) + C(14, 5) + ... + C(8, 5)).
Step-by-step explanation:
To find the number of solutions to the equation x₁ + x₂ + x₃ + x₄ + x₅ + x₆ = 25, we can use a technique called stars and bars. In this case, there are 25 stars (representing the value 1) and 5 bars (to separate the 6 variables).
1) There are no other restrictions:
The number of solutions is C(25 + 6 - 1, 6 - 1) = C(30, 5) = 142,506.
2) x₁ ≥ 3 for i = 1, 2, 3, 4, 5, 6:
First, we subtract 3 from each variable: y₁ = x₁ - 3, y₂ = x₂ - 3, ..., y₆ = x₆ - 3. Then, we can use the same approach as in part 1:
The number of solutions is C(25 - 6(3) + 6 - 1, 6 - 1) = C(4, 5) = 0. There are no solutions.
3) 3 ≤ x₁ ≤ 10:
Let's consider the number of solutions for each possible value of x₁. For x₁ = 3, there are C(22, 5) solutions. For x₁ = 4, there are C(21, 5) solutions. Continuing this pattern, for x₁ = 10, there are C(15, 5) solutions.
The total number of solutions is (C(22, 5) + C(21, 5) + ... + C(15, 5)).
4) 3 ≤ x₁ ≤ 10 and 2 ≤ x₂ ≤ 7:
We can use the same approach as in part 3, but this time we need to consider the overlapping range of x₁ and x₂. Let's consider each possible value of x₁ and x₂ separately and find the number of solutions:
The total number of solutions is (C(15, 5) + C(14, 5) + ... + C(8, 5)).