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MinusTheBear
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Homework Statement
How many nonnegative integer solutions does the following equation has x + y + z = 15 if x ≥ 3, y ≥ 2, and 1 ≤ z ≤ 3?
Homework Equations
The Attempt at a Solution
Part A:
Since x ≥ 3, y ≥ 2
let a = x - 3 and b = y - 2
a + b + z = 15 - 3 - 2
a + b + z = 10
Therefore,
n = 10
r = 3
via r-combination with repetition formula there are C(10 + 3 - 1, 10) ways to complete part A. That is C(12,10) = 66 ways.
Part B:
From part A we have
a + b + z = 10
Since 1 ≤ z ≤ 3 subtracting 1 from both sides we get 0 ≤ z ≤ 2 that is z ≤ 2, we must also subtract 1 from the equation
a + b + z = 10 - 1 = 9
Since z is at most 2, we can use z compliment and subtract it from part a.
The complement of z is z ≥ 3, let c = z - 3
a + b + c = 9 - 3
a + b + c = 6
Therefore,
n = 6
r = 3
via r-combination with repetition formula there are C(6+3-1, 6) ways to complete part B. That is C(8,6) = 28 ways.
Since we took the compliment in Part B, we must subtract it from Part A. So,
66 - 28 = 38.
MY ANSWER: 38
SOLUTION: 27.
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