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 \(2 \cdot 3 \cdot 4 - 1 = \boxed{23}.\)