Using C(n, r) to denote a binomial coefficient, how many integer solutions…
2012
Using C(n, r) to denote a binomial coefficient, how many integer solutions does x1 + x2 + x3 = 11 have when x1 ≥ 1, x2 ≥ 2, and x3 ≥ 3?
Answer: D. C(7, 5) — ConceptFor nonnegative integer variables in y1 + y2 + ⋯ + yk = n, the stars-and-bars principle gives C(n + k − 1, k − 1). Lower bounds are handled by…
- A.
C(7, 11)
- B.
C(11, 3)
- C.
C(14, 11)
- D.
C(7, 5)
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Correct answer: D
Concept
For nonnegative integer variables in y1 + y2 + ⋯ + yk = n, the stars-and-bars principle gives C(n + k − 1, k − 1). Lower bounds are handled by subtracting each variable’s minimum value before applying the formula.
Application
Define y1 = x1 − 1, y2 = x2 − 2, and y3 = x3 − 3. Then y1, y2, and y3 are nonnegative integers.
Substitute the definitions: (y1 + 1) + (y2 + 2) + (y3 + 3) = 11, so y1 + y2 + y3 = 5.
Apply stars and bars with n = 5 and k = 3: C(5 + 3 − 1, 3 − 1) = C(7, 2) = C(7, 5) = 21.
Cross-check
Fixing y1 at 0, 1, 2, 3, 4, or 5 leaves respectively 6, 5, 4, 3, 2, or 1 nonnegative pairs (y2, y3). Their total is 6 + 5 + 4 + 3 + 2 + 1 = 21. Therefore, the required number of solutions is C(7, 5).