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).
Important Questions from Permutation and Combination
If Quantity A is the number of ways to assign a number from 1 to 5 without repetition to each of four people, and Quantity B is the number of ways to assign a number from 1 to 5 without repetition to each of 5 people, then which of the following statements is correct with respect to Quantities A and B?
Which of the following muscles regulates the exit of food from the stomach into the small intestine?
There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
There are $15$ distinct points on a plain sheet of paper. If $4$ of these points are collinear, find the maximum number of triangles that can be drawn using these points.
