- #1

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:

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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