We are asked to determine how many ways a person can make an 80-cent purchase using exactly five nickels, three dimes, and two quarters, considering all coins within a particular value to be identical.

Step 1: Write down the values of the coins.

• Each nickel is worth 5 cents.
• Each dime is worth 10 cents.
• Each quarter is worth 25 cents.

Step 2: Define variables.

Let:

• $x$ be the number of nickels used.
• $y$ be the number of dimes used.
• $z$ be the number of quarters used.

We want the total value of the coins used to be 80 cents: $5x + 10y + 25z = 80$ Additionally, the constraints are:

• $0 \leq x \leq 5$ (you have at most 5 nickels),
• $0 \leq y \leq 3$ (you have at most 3 dimes),
• $0 \leq z \leq 2$ (you have at most 2 quarters).

Step 3: Solve for possible combinations of $x$, $y$, and $z$.

We now systematically check for different values of $z$ (quarters) and solve for the corresponding $x$ (nickels) and $y$ (dimes).

Case 1: $z = 2$

If you use 2 quarters, the value from the quarters is: $25 \times 2 = 50 \text{ cents}.$ This leaves: $80 - 50 = 30 \text{ cents}.$ Now solve $5x + 10y = 30$:

• If $y = 0$, $5x = 30 \Rightarrow x = 6$ (not possible because $x \leq 5$).
• If $y = 1$, $5x + 10 = 30 \Rightarrow 5x = 20 \Rightarrow x = 4$ (valid).
• If $y = 2$, $5x + 20 = 30 \Rightarrow 5x = 10 \Rightarrow x = 2$ (valid).
• If $y = 3$, $5x + 30 = 30 \Rightarrow 5x = 0 \Rightarrow x = 0$ (valid).

Thus, for $z = 2$, the valid combinations are: $(4, 1, 2), (2, 2, 2), (0, 3, 2).$

Case 2: $z = 1$

If you use 1 quarter, the value from the quarter is: $25 \times 1 = 25 \text{ cents}.$ This leaves: $80 - 25 = 55 \text{ cents}.$ Now solve ( 5x +