Difference between revisions of "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?"
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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> | 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, | + | 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. |
Latest revision as of 23:18, 1 January 2024
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 
possible combinations. But we must remove the combination where Steve does not use any coins, leaving us with 
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.
