Introduction / Context:
This is a classic problem of finding the number of positive integer solutions to a linear equation in several variables. We are given a + b + c = 21 with the condition that a, b and c are all positive integers, and we need to count how many ordered triples satisfy this equation.
Given Data / Assumptions:
- a, b and c are positive integers (strictly greater than zero).
- a + b + c = 21.
- Different orders of the same numbers count as different solutions.
Concept / Approach:
This is a stars and bars problem. For positive integer solutions, the standard formula says that the number of solutions to a1 + a2 + ... + ak = n, with each ai greater than zero, is C(n - 1, k - 1). Here k = 3 and n = 21.
Step-by-Step Solution:
Step 1: Recognise that a, b and c are all strictly positive, which matches the positive integer stars and bars formula.
Step 2: Use the formula for the number of positive integer solutions: C(n - 1, k - 1).
Step 3: Substitute n = 21 and k = 3.
Step 4: Compute C(21 - 1, 3 - 1) = C(20, 2).
Step 5: Evaluate C(20, 2) = 20 * 19 / (2 * 1) = 190.
Verification / Alternative check:
You can transform variables by letting a = x + 1, b = y + 1 and c = z + 1 where x, y and z are non negative integers. Then x + y + z = 21 - 3 = 18. The number of non negative integer solutions of x + y + z = 18 is C(18 + 3 - 1, 3 - 1) = C(20, 2) = 190, confirming the result.
Why Other Options Are Wrong:
109, 901 and 910 do not match the correct combinatorial formula for three positive integers summing to 21. They are not values of C(20, 2) or any simple binomial coefficient for this setup.
Common Pitfalls:
A common error is to use the formula for non negative solutions directly on 21, which would give C(21 + 3 - 1, 3 - 1) and an incorrect higher value. Another mistake is to forget that order matters, but in this formula ordered triples are already counted correctly.
Final Answer:
The total number of ordered positive integer triples (a, b, c) such that a + b + c = 21 is
190.
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