Finding Positive Integer Solutions for Linear Equations

The problem asks for the number of different solutions to the equation $\mathtt{x} + \mathtt{y} + \mathtt{z} = 12$, where each of $\mathtt{x}$, $\mathtt{y}$, and $\mathtt{z}$ must be a positive integer. This is a common type of combinatorics problem that can be solved using the stars and bars method.

The stars and bars method is typically used to find the number of non-negative integer solutions to an equation of the form $\mathtt{x_1} + \mathtt{x_2} + \dots + \mathtt{x_k} = \mathtt{N}$. In this problem, however, the variables $\mathtt{x}$, $\mathtt{y}$, and $\mathtt{z}$ must be positive integers ($\ge 1$), not non-negative integers ($\ge 0$).

Transforming to Non-Negative Integers

To handle the positive integer constraint, we can transform the variables. Since $\mathtt{x}$, $\mathtt{y}$, $\mathtt{z}$ must be at least 1, we can define new variables that are non-negative. Let:

• $\mathtt{x} = \mathtt{x'} + 1$
• $\mathtt{y} = \mathtt{y'} + 1$
• $\mathtt{z} = \mathtt{z'} + 1$

where $\mathtt{x'} \ge 0$, $\mathtt{y'} \ge 0$, and $\mathtt{z'} \ge 0$ are non-negative integers. If $\mathtt{x'}$, $\mathtt{y'}$, $\mathtt{z'}$ are any non-negative integers, then $\mathtt{x}$, $\mathtt{y}$, $\mathtt{z}$ will automatically be positive integers.

Solving the Transformed Equation

Substitute these new expressions for $\mathtt{x}$, $\mathtt{y}$, and $\mathtt{z}$ into the original equation:

$(\mathtt{x'} + 1) + (\mathtt{y'} + 1) + (\mathtt{z'} + 1) = 12$

Simplify the equation:

$\mathtt{x'} + \mathtt{y'} + \mathtt{z'} + 3 = 12$

Subtract 3 from both sides:

$\mathtt{x'} + \mathtt{y'} + \mathtt{z'} = 12 - 3$

$\mathtt{x'} + \mathtt{y'} + \mathtt{z'} = 9$

Now, the problem is equivalent to finding the number of non-negative integer solutions to the equation $\mathtt{x'} + \mathtt{y'} + \mathtt{z'} = 9$.

Applying the Stars and Bars Method

The stars and bars formula for finding the number of non-negative integer solutions to $\mathtt{n_1} + \mathtt{n_2} + \dots + \mathtt{n_k} = \mathtt{N}$ is given by:

$\binom{\mathtt{N} + \mathtt{k} - 1}{\mathtt{k} - 1}$ or $\binom{\mathtt{N} + \mathtt{k} - 1}{\mathtt{N}}$

In our transformed equation $\mathtt{x'} + \mathtt{y'} + \mathtt{z'} = 9$:

• $\mathtt{N} = 9$ (the sum)
• $\mathtt{k} = 3$ (the number of variables: $\mathtt{x'}$, $\mathtt{y'}$, $\mathtt{z'}$)

Using the formula with $\mathtt{N}=9$ and $\mathtt{k}=3$, the number of solutions is:

$\binom{9 + 3 - 1}{3 - 1} = \binom{11}{2}$

Now, we calculate the value of the combination $\binom{11}{2}$:

$\binom{11}{2} = \frac{11!}{2!(11-2)!} = \frac{11!}{2!9!} = \frac{11 \times 10 \times 9!}{2 \times 1 \times 9!} = \frac{11 \times 10}{2} = \frac{110}{2} = 55$

Alternatively, using $\binom{\mathtt{N} + \mathtt{k} - 1}{\mathtt{N}}$:

$\binom{9 + 3 - 1}{9} = \binom{11}{9}$

$\binom{11}{9} = \frac{11!}{9!(11-9)!} = \frac{11!}{9!2!} = \frac{11 \times 10 \times 9!}{9! \times 2 \times 1} = \frac{11 \times 10}{2} = \frac{110}{2} = 55$

Both methods give the same result.

Therefore, there are 55 non-negative integer solutions for $\mathtt{x'} + \mathtt{y'} + \mathtt{z'} = 9$. Each of these solutions corresponds uniquely to a positive integer solution for the original equation $\mathtt{x} + \mathtt{y} + \mathtt{z} = 12$.

Conclusion on the Number of Solutions

The number of different positive integer solutions of the equation $\mathtt{x} + \mathtt{y} + \mathtt{z} = 12$ is 55.

Revision Table: Key Steps for Positive Integer Solutions

Additional Information: Stars and Bars Method

The stars and bars method is a visual technique used in combinatorics. To find the number of non-negative integer solutions to $\mathtt{n_1} + \mathtt{n_2} + \dots + \mathtt{n_k} = \mathtt{N}$, imagine $\mathtt{N}$ identical items ("stars") that you want to distribute into $\mathtt{k}$ distinct bins. This is equivalent to placing $\mathtt{k}-1$ "bars" among the $\mathtt{N}$ stars.

Consider the equation $\mathtt{x'} + \mathtt{y'} + \mathtt{z'} = 9$ with non-negative integers. We have 9 "stars" (units summing up to 9) and we need to divide them among 3 variables ($\mathtt{x'}$, $\mathtt{y'}$, $\mathtt{z'}$). We need $\mathtt{k}-1 = 3-1 = 2$ "bars" to create three sections for the variables' values.

For example, if we have 9 stars ($\ast\ast\ast\ast\ast\ast\ast\ast\ast$) and 2 bars ($\vert$), a possible arrangement could be:

$\ast\ast\ast \vert \ast\ast\ast\ast \vert \ast\ast$

This corresponds to $\mathtt{x'} = 3$, $\mathtt{y'} = 4$, $\mathtt{z'} = 2$, which is a non-negative integer solution to $\mathtt{x'} + \mathtt{y'} + \mathtt{z'} = 9$. The number of such arrangements is the number of ways to choose the positions of the 2 bars among the total $\mathtt{N} + \mathtt{k} - 1 = 9 + 3 - 1 = 11$ positions (9 stars and 2 bars). This is given by $\binom{11}{2}$.

This method is versatile and can be adapted for other constraints, such as lower bounds other than 1 or 0, or upper bounds (though upper bounds make it more complex, often requiring the principle of inclusion-exclusion).