View source for Steve has one quarter, two nickels and three pennies. Assuming no items are free, for how many different-priced items could Steve individually pay for with exact change?

Steve can use no quarters or one quarter, for two possibilities.

Steve can use 0, 1, or 2 nickels, for three possibilities.

And Steve can use 0, 1, 2, or 3 pennies, for four possibilities. That gives <math>2 \cdot 3 \cdot 4 = 24</math> possible combinations. But we must remove the combination where Steve does not use any coins, leaving us with <math>24 - 1 = \boxed{23}.</math>

Note that this only works because each obtainable value can only be obtained in 1 way. In other words, if there were 5 pennies instead of 3, there would be 2 ways to buy something that costs 5 cents: 1 nickel or 5 pennies.