Solution
We shall generalize this problem. Let one gold coin worth A silver, and one silver coin worth B bronze coins. The price of the item is thus 3AB bronze coins. We will next consider the number of gold coins used to pay.
If Ali uses 3 gold coins, there is only one way to pay, that is, using 3 gold coins.
If Ali uses 2 gold coins, then he must use a combination of silver and bronze to pay for the remaining price that's equivalent to AB bronze coins. He could use 0,1,cdots, A silver coins to pay for this. Once he decides the number of silver coins used, the remaining price must be paid all by bronze coins. So there are (A+1) ways to pay with exacty 2 gold coins.
Similarly, there are (2A+1) ways to pay with exactly 1 gold coin, and (3A+1) ways to pay without any gold coins.
In total, there are 1 + (A+1) + (2A+1) + (3A+1) = 6A+1 = 2008 ways to pay.
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