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How Many Solutions Are There If x1 + x2 + x3 = 10? A Complete Guide
Finding the number of non-negative integer solutions to the equation x1 + x2 + x3 = 10 might seem daunting at first, but with the right approach, it's easily solvable. This comprehensive guide will walk you through the solution using the stars and bars method, a powerful combinatorial technique. We'll also explore variations and extensions of this problem to solidify your understanding.
Understanding the Stars and Bars Method
The stars and bars method is a classic technique for solving problems like this. Imagine you have 10 identical items (stars) to distribute among 3 distinct containers (x1, x2, x3). The number of ways to do this is equivalent to the number of solutions to our equation.
To visualize this, think of placing bars between the stars to separate them into groups. For example, if we have 10 stars and we want to divide them into 3 groups, we need 2 bars.
*****|*****| This represents x1 = 5, x2 = 5, x3 = 0.
The total number of positions is 10 + 2 = 12. We need to choose 2 positions for the bars out of these 12 positions. This is a combination problem, and the formula is:
C(n + k - 1, k - 1) = C(n + k - 1, n)
where:
- n is the sum (10 in our case)
- k is the number of variables (3 in our case)
Calculating the Number of Solutions
Using the formula above, we plug in our values:
C(10 + 3 - 1, 3 - 1) = C(12, 2)
This represents the number of ways to choose 2 positions for the bars among 12 total positions. Calculating this combination:
C(12, 2) = 12! / (2! * 10!) = (12 * 11) / (2 * 1) = 66
Therefore, there are 66 non-negative integer solutions to the equation x1 + x2 + x3 = 10.
Extending the Concept: Variations and Constraints
This method can be adapted to handle various constraints:
Positive Integer Solutions:
If we require x1, x2, and x3 to be positive integers (greater than 0), we can use a substitution. Let y1 = x1 - 1, y2 = x2 - 1, and y3 = x3 - 1. Our equation becomes:
y1 + 1 + y2 + 1 + y3 + 1 = 10
y1 + y2 + y3 = 7
Now we use the stars and bars method again with n = 7 and k = 3:
C(7 + 3 - 1, 3 - 1) = C(9, 2) = 36
There are 36 solutions where x1, x2, x3 are positive integers.
Solutions with Upper Bounds:
If we add upper bounds to the variables (e.g., x1 β€ 5), we would need to use the Principle of Inclusion-Exclusion to subtract the cases where the constraints are violated. This becomes more complex but is still manageable with systematic counting.
Conclusion: Mastering Combinatorial Techniques
Understanding the stars and bars method is crucial for solving a wide range of combinatorial problems. This technique provides a clear and efficient way to determine the number of solutions to linear equations with non-negative integer constraints. By mastering this method and its variations, you'll be well-equipped to tackle more advanced problems in combinatorics and related fields. Remember to consider the constraints carefully, and adapt your approach accordingly.
