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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?"

(Created page with "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...")
 
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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>
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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.
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{{delete|long title}}

Latest revision as of 15:28, 30 October 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 $2 \cdot 3 \cdot 4 = 24$ possible combinations. But we must remove the combination where Steve does not use any coins, leaving us with $24 - 1 = \boxed{23}.$

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.

This article has been proposed for deletion. The reason given is: long title.

Sysops: Before deleting this article, please check the article discussion pages and history.

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