How many solutions are there for the equation 𝑥 + 𝑦 + 𝑧 + 𝑢 = 29 subject…
2015
How many solutions are there for the equation 𝑥 + 𝑦 + 𝑧 + 𝑢 = 29 subject to the constraints that 𝑥 ≥ 1, 𝑦 ≥ 2, 𝑧 ≥ 3 and 𝑢 ≥ 0?
- A.
4960
- B.
2600
- C.
23751
- D.
8855
Attempted by 37 students.
Show answer & explanation
Correct answer: B
Answer: 2600
Key idea: convert to nonnegative variables by subtracting the minimum required values, then use stars and bars to count solutions.
Let x' = x - 1, y' = y - 2, z' = z - 3, and u' = u. Then x', y', z', u' are nonnegative integers.
Substitute into the equation: x' + y' + z' + u' = 29 - (1 + 2 + 3) = 23.
The number of nonnegative integer solutions for four variables summing to 23 is C(23 + 4 - 1, 4 - 1) = C(26, 3).
Compute C(26, 3) = 26 * 25 * 24 / 6 = 2600.