The number of different solutions of the equations x + y + z = 12, where each of x,y and z is a positive integer, is
55
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
| Step | Description | Equation Form |
|---|---|---|
| 1 | Identify the equation and constraint (positive integers). | \(\mathtt{x} + \mathtt{y} + \mathtt{z} = 12\) (\(\mathtt{x, y, z} \ge 1\)) |
| 2 | Transform variables to be non-negative (\(\ge 0\)). | Let \(\mathtt{x} = \mathtt{x'} + 1\), \(\mathtt{y} = \mathtt{y'} + 1\), \(\mathtt{z} = \mathtt{z'} + 1\) |
| 3 | Substitute and simplify the equation. | \(\mathtt{x'} + \mathtt{y'} + \mathtt{z'} = 9\) (\(\mathtt{x', y', z'} \ge 0\)) |
| 4 | Apply Stars and Bars formula \(\binom{\mathtt{N} + \mathtt{k} - 1}{\mathtt{k} - 1}\). | \(\mathtt{N}=9\), \(\mathtt{k}=3\). Formula: \(\binom{9+3-1}{3-1} = \binom{11}{2}\) |
| 5 | Calculate the combination. | \(\binom{11}{2} = 55\) |
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).
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