Key Concepts and Formulas

• Stars and Bars: The number of non-negative integer solutions to $x_1 + x_2 + \dots + x_n = r$ is given by ${n+r-1 \choose n-1} = {n+r-1 \choose r}$.
• Variable Transformation: If $x \ge a$, substitute $x' = x - a$ to obtain a non-negative variable $x' \ge 0$.
• Inclusion-Exclusion Principle: Used to correct for overcounting when dealing with multiple constraints. In simple cases, it involves subtracting solutions that violate constraints.

Step-by-Step Solution

Step 1: Define the variables and the equation

We are looking for 3-digit numbers whose digits sum to 10. Let the digits be $x, y, z$, where $x$ is the hundreds digit, $y$ is the tens digit, and $z$ is the units digit. We want to find the number of integer solutions to the equation:

$x + y + z = 10$

subject to the constraints:

• $1 \le x \le 9$ (since $x$ is the hundreds digit)
• $0 \le y \le 9$
• $0 \le z \le 9$

Step 2: Transform the equation for non-negative solutions

Since $x \ge 1$, we introduce a new variable $x' = x - 1$, so $x = x' + 1$ and $x' \ge 0$. Substituting into the equation, we get:

$(x' + 1) + y + z = 10$

$x' + y + z = 9$

Now we have the equation $x' + y + z = 9$ with the constraints $x' \ge 0$, $y \ge 0$, and $z \ge 0$. Also, since $x=x'+1$, $x' \le 8$.

Step 3: Apply Stars and Bars to find the initial number of solutions

We can apply the Stars and Bars formula to find the number of non-negative integer solutions to $x' + y + z = 9$. Here, $n = 3$ (number of variables) and $r = 9$ (the sum). The number of solutions is:

${n+r-1 \choose n-1} = {3+9-1 \choose 3-1} = {11 \choose 2} = \frac{11 \times 10}{2 \times 1} = 55$

This gives us an initial count of 55 solutions.

Step 4: Account for the upper bound violations

We need to consider the constraints $x \le 9$, $y \le 9$, and $z \le 9$, which translate to $x' \le 8$, $y \le 9$, and $z \le 9$. We will subtract the cases where these conditions are violated.

• Case 1: $x > 9$ (or $x' > 8$) If $x' > 8$, then $x' \ge 9$. Let $x'' = x' - 9$, so $x' = x'' + 9$ and $x'' \ge 0$. Substituting into $x' + y + z = 9$:

  $(x'' + 9) + y + z = 9$

  $x'' + y + z = 0$

  The only non-negative integer solution is $x'' = 0, y = 0, z = 0$. This corresponds to $x' = 9, y = 0, z = 0$, which means $x = 10, y = 0, z = 0$. This is one solution where $x > 9$.

• Case 2: $y > 9$ If $y > 9$, then $y \ge 10$. Let $y' = y - 10$, so $y = y' + 10$ and $y' \ge 0$. Substituting into $x' + y + z = 9$:

  $x' + (y' + 10) + z = 9$

  $x' + y' + z = -1$

  There are no non-negative integer solutions to this equation.

• Case 3: $z > 9$ If $z > 9$, then $z \ge 10$. Let $z' = z - 10$, so $z = z' + 10$ and $z' \ge 0$. Substituting into $x' + y + z = 9$:

  $x' + y + (z' + 10) = 9$

  $x' + y + z' = -1$

  There are no non-negative integer solutions to this equation.

Therefore, only one solution violates the constraints.

Step 5: Calculate the final answer

The number of valid solutions is the initial number of solutions minus the number of violations:

$55 - 1 = 54$

Therefore, there are 54 such 3-digit numbers.

Common Mistakes & Tips

• Always remember to transform variables to ensure they are non-negative before applying the Stars and Bars formula.
• Carefully check for upper bound violations and use the Inclusion-Exclusion Principle to correct for overcounting, if necessary. In this problem, only one violation needed to be removed.
• When applying the Inclusion-Exclusion principle, consider all single-variable violations first, then combinations of two variables, and so on.

Summary

We used the Stars and Bars method to find the number of solutions to $x+y+z=10$ with the given constraints on $x, y, z$. We first transformed the equation to have non-negative variables. Then, we applied the Stars and Bars formula and accounted for the upper bound violations. The final answer is 54.

The final answer is \boxed{54}.