+199 1 item for 1 penny
1 item for 2 pennies
1 item for 3 pennies
1 item for 5 cents
1 item for 6 cents
1 item for 7 cents
1 item for 8 cents
1 item for 10 cents
1 item for 11 cents
1 item for 12 cents
1 item for 13 cents
1 item for 25 cents
1 item for 26 cents
1 item for 27 cents
1 item for 28 cents
1 item for 30 cents
1 item for 31 cents
1 item for 32 cents
1 item for 33 cents
1 item for 35 cents
1 item for 36 cents
1 item for 37 cents
1 item for 38 cents.
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Total 23 different items for exact change.
+42 Note that there is no way to get the same amount from two different combinations of coins. Thus, it suffices to find the nmber of ways to choose some nonzero number of coins from the coins Steve has where coins of the same value are considered identical.
We can have zero or one quarter, so there are two ways to choose how many quarters.
We can have zero, one, or two nickels, so there are three ways to choose how many nickels.
We can have any where from zero to three pennies, so there are four ways to choose the pennies.
However, note that we included the case when we choose no coins at all, which is not allowed, so we subtract \(1\) from the product of \(2,3,\)and \(4\), to get \(2 \cdot 3 \cdot 4 - 1 = \boxed{23}.\)
