Calculus 2 Counting Pie

Question

Consider the equation x1+x2+x3+x4=16. How many solutions are there with 2≤xi≤6 for all i∈{1,2,3,4}

Answer 1

I need to find the number of non-negative integer solutions to x₁ + x₂ + x₃ + x₄ = 16 where 2 ≤ xᵢ ≤ 6 for all i.

Method: Change of Variables

Let me substitute yᵢ = xᵢ - 2 for each i. This transforms the constraints:

1. Since 2 ≤ xᵢ ≤ 6, we get 0 ≤ yᵢ ≤ 4

The equation becomes: (y₁ + 2) + (y₂ + 2) + (y₃ + 2) + (y₄ + 2) = 16

y₁ + y₂ + y₃ + y₄ + 8 = 16

y₁ + y₂ + y₃ + y₄ = 8 where 0 ≤ yᵢ ≤ 4

Using Inclusion-Exclusion

First, without the upper bound constraint, the number of non-negative integer solutions to y₁ + y₂ + y₃ + y₄ = 8 is: (8+4−14−1)=(113)=165(4−18+4−1​)=(311​)=165

Now I need to subtract cases where at least one yᵢ > 4 (i.e., yᵢ ≥ 5).

Let Aᵢ = set of solutions where yᵢ ≥ 5

For |A₁|: If y₁ ≥ 5, let y₁ = z₁ + 5 where z₁ ≥ 0. Then z₁ + 5 + y₂ + y₃ + y₄ = 8, so z₁ + y₂ + y₃ + y₄ = 3

Number of solutions: $\binom{3+4-1}{4-1} = \binom{6}{3} = 20$

By symmetry: |A₁| = |A₂| = |A₃| = |A₄| = 20

For |A₁ ∩ A₂|: If y₁, y₂ ≥ 5, then y₁ + y₂ ≥ 10 > 8. This is impossible.

Therefore, all higher-order intersections are also empty.

Final Answer

By inclusion-exclusion: Answer=165−4(20)=165−80=85Answer=165−4(20)=165−80=85​