The problem is asking for the number of distinct solutions to the equation

$x_1 + x_2 + x_3 + x_4 = 100$

with the constraints:

• $x_1 \in \{0, 1, 2, \dots, 10\}$ (meaning $x_1$ is bounded between 0 and 10),
• $x_2, x_3, x_4 \in \{0, 1, 2, \dots\}$ (meaning $x_2, x_3, x_4$ are non-negative integers).

This is a problem involving integer partitioning with constraints. Let’s break it down step by step:

Step 1: Consider the unrestricted case.

If there were no restrictions on $x_1$ (so all $x_1, x_2, x_3, x_4$ were non-negative integers), the number of solutions to this equation would be a standard stars-and-bars problem. The number of non-negative integer solutions to:

$x_1 + x_2 + x_3 + x_4 = 100$

is given by the stars-and-bars formula, which counts the number of ways to distribute 100 units among 4 variables:

$\binom{100 + 4 - 1}{4 - 1} = \binom{103}{3}.$

Step 2: Apply the constraint on $x_1$.

We need to account for the fact that $x_1$ is restricted to the set $\{0, 1, 2, \dots, 10\}$. To handle this, we use the principle of inclusion-exclusion.

Unrestricted case for $x_1$:

If we disregard the constraint on $x_1$, the number of solutions is:

$\binom{103}{3}.$

Subtract the invalid cases:

Now, subtract the cases where $x_1 > 10$. If $x_1 > 10$, let $x_1' = x_1 - 11$, where $x_1' \geq 0$. In terms of $x_1'$, the equation becomes:

$x_1' + x_2 + x_3 + x_4 = 89$

The number of non-negative integer solutions to this equation is:

$\binom{89 + 4 - 1}{4 - 1} = \binom{92}{3}.$

Step 3: Conclusion.

Using inclusion-exclusion, the total number of solutions is:

$\binom{103}{3} - \binom{92}{3}.$

Now, we compute the binomial coefficients:

$\binom{103}{3} = \frac{103 \times 102 \times 101}{3 \times 2 \times 1} = 176851,$

$\binom{92}{3} = \frac{92 \times 91 \times 90}{3 \times 2 \times 1} = 125580.$

Thus, the total number of distinct solutions is:

$176851 - 125580 = 51271.$

Final Answer:

There are 51,271 distinct solutions to the equation.

Would you like to go over any details, or have any questions?

Here are 5 related questions to explore further:

1. What is the stars-and-bars method for solving combinatorial problems?
2. How does inclusion-exclusion work in combinatorics?
3. Can we generalize this problem to more variables or different ranges for the variables?
4. How would the problem change if more variables were bounded like $x_1$?
5. What are other methods for counting solutions to Diophantine equations?

Tip: When solving combinatorial problems with restrictions, inclusion-exclusion is a powerful tool to account for overcounting!