To solve this problem, we need to determine how many ways a person can make an 80-cent purchase using five nickels, three dimes, and two quarters. The approach is to look for combinations of these coins that sum up to exactly 80 cents, keeping in mind that the total number of coins used in each denomination cannot exceed the number available.

Step 1: Define variables for the number of each type of coin.

Let:

• $x_1$ = number of nickels used (each worth 5 cents),
• $x_2$ = number of dimes used (each worth 10 cents),
• $x_3$ = number of quarters used (each worth 25 cents).

We need to find all non-negative integer solutions to the equation: $5x_1 + 10x_2 + 25x_3 = 80$ subject to the following constraints: $0 \leq x_1 \leq 5, \quad 0 \leq x_2 \leq 3, \quad 0 \leq x_3 \leq 2$

Step 2: Simplify the equation.

Dividing the entire equation by 5: $x_1 + 2x_2 + 5x_3 = 16$ Now, we need to find all combinations of $x_1$, $x_2$, and $x_3$ that satisfy this equation while respecting the constraints.

Step 3: Try different values for $x_3$.

We will check all possible values for $x_3$ and solve for $x_1 + 2x_2 = 16 - 5x_3$.

• For $x_3 = 0$: $x_1 + 2x_2 = 16$ Possible pairs $(x_1, x_2)$ are:

  • $x_1 = 16, x_2 = 0$ (invalid since $x_1 > 5$)
  • $x_1 = 14, x_2 = 1$ (invalid since $x_1 > 5$)
  • $x_1 = 12, x_2 = 2$ (invalid since $x_1 > 5$)
  • $x_1 = 10, x_2 = 3$ (invalid since $x_1 > 5$)

  No valid solutions for $x_3 = 0$.

• For $x_3 = 1$: $x_1 + 2x_2 = 11$ Possible pairs $(x_1, x_2)$ are:

  • $x_1 = 11, x_2 = 0$ (invalid)
  • $x_1 = 9, x_2 = 1$ (invalid)
  • $x_1 = 7, x_2 = 2$ (invalid)
  • $x_1 = 5, x_2 = 3$ (valid)

  One valid solution: $(x_1, x_2, x_3) = (5, 3, 1)$.

• For $x_3 = 2$: $x_1 + 2x_2 = 6$ Possible pairs $(x_1, x_2)$ are:

  • $x_1 = 6, x_2 = 0$ (invalid)
  • $x_1 = 4, x_2 = 1$ (valid)
  • $x_1 = 2, x_2 = 2$ (valid)
  • $x_1 = 0, x_2 = 3$ (valid)

  Three valid solutions: $(4, 1, 2), (2, 2, 2), (0, 3, 2)$.

Step 4: Count the valid solutions.

The valid solutions are:

• $(x_1, x_2, x_3) = (5, 3, 1)$,
• $(x_1, x_2, x_3) = (4, 1, 2)$,
• $(x_1, x_2, x_3) = (2, 2, 2)$,
• $(x_1, x_2, x_3) = (0, 3, 2)$.

Thus, there are 4 ways to make the 80-cent purchase.

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Do you need more details or have any questions?

Here are 5 related questions to expand on this:

1. How many ways can a person make a 60-cent purchase with the same number of coins?
2. What if there are additional pennies available for making the exact change?
3. How can this approach be extended to solve similar problems with different total amounts?
4. What are some strategies for organizing and solving coin combination problems more efficiently?
5. How can generating functions be used to solve coin change problems in combinatorics?

Tip: When solving coin combination problems, simplify the equation by factoring out common values to reduce complexity.