First, for simplicity, let's make all the amounts of money into integers by considering them all in cents. For example, $ \ $5.43$ becomes 543. Let the purchase price be $A=A_1A_2A_3$ and the amount of change be $B_1B_2B_3$ where $A_1$ represents the first digit of $A$ , $B_1$ represents the first digit of $B$ , $A_2$ represents the second digit of $A$ , etc. We know that $A+B=1000$ , and we can conclude that $A_1+B_1=9$ because if $A_1+B_1<9$ then $A+B<1000$ and if $A_1+B_1=10$ then $A_2=B_2=A_3=B_3=0$ , but then the only way that B can be a rearrangement of the digits of A is if $A_1=B_1=5$ , which means $A=B=500$ , but the problem states that the price and the amount of change are different. Since 9 is odd, we can also conclude that $A_1$ and $B_1$ are distinct, which, using the fact that $A$ 's digits can be rearranged to get B's digits, implies that $A_1=B_2$ or $A_1=B_3$ and $B_1=A_2$ or $B_1=A_3$ . We can also observe that A and B have the same remainder when divided by 9 because the remainder when $n$ is divided by 9 is equal to the remainder when the sum of the digits of $n$ is divided by 9 for all $n$ and the sum of the digits of A is obviously equal to the sum of the digits of B. Since the remainder when 1000 is divided by 9 is 1, we can in fact conclude that the remainder when A and B are divided by 9 (and when the sum of their digits is divided by 9) is 5. Keeping in mind that two of the digits of $A$ are $A_1$ and $B_1$ and that $A_1+B_1=9$ , we can conclude that the other digit is 5, which is the only digit that would result in the sum having a remainder of 5 when divided by 9. By similar logic we can conclude that 5 is also one of the digits of $B$ . A little thought makes it clear that at least one of these 5's appears as the last digit in its number (that is, $A_3=5$ or $B_3=5$ ) since if neither of them appears as the last digit in a number, then $A_1=B_3$ and $B_1=A_3$ and $A_3+B_3=9\Rightarrow A+B$ ends in a 9, which is a contradiction. But if $A_3=5$ then the only way for the sum of $A$ and $B$ to end in a 0 is for $B_3=5$ , so we can conclude that $A_3=B_3=5$ , $A_1=B_2$ , and $A_2=B_1$ . So once we have picked a value for $A_1$ , the other 5 digits are all determined. Since both amounts are greater than a dollar, we know that $A_1$ can be any number between 1 and 8 for a total of 8 possible prices (and thus 8 possible amounts of change). To double check, we can work out $A$ and $B$ for each value of $A_1$ and reconvert them to dollars to make sure that the price and the amount of change satisfy the given conditions. $A_1=1\Rightarrow A=\$1.85, B=\$8.15$ ; $A_1=2\Rightarrow A=\$2.75, B=\$7.25$ ; $A_1=3\Rightarrow A=\$3.65, B=\$6.35$ ; $A_1=4\Rightarrow A=\$4.55, B=\$5.45$ ; $A_1=5\Rightarrow A=\$5.45, B=\$4.55$ ; $A_1=6\Rightarrow A=\$6.35, B=\$3.65$ ; $A_1=7\Rightarrow A=\$7.25, B=\$2.75$ ; and finally $A_1=8\Rightarrow A=\$8.15, B=\$1.85$. This confirms that there are $\boxed{8}$ possible amounts of change.