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# i have 3 quarters, 2 dimes, a nickle & 2 pennies.?

So how many different amounts of money can i make using some or all of these coins?

The answer is 62 but what are the 62 combinations ie. 1q+1n+1d,

1q+1n+2d,

### 2 Answers

- Jenna KLv 61 decade agoFavorite Answer
There are 72 possible combinations of coins, with 62 unique amounts of money. And it's pointless to list all 72 combinations. You "count" them by multiplying the possibilities.

You can use 0, 1, 2, or 3 quarters, so there are 4 possibilities.

For each of the 4 possibilities of quarters, you can use 0, 1, or 2 dimes. So there are 3 possibilities for the number of dimes, and 4*3=12 possibilities for the number of dimes and quarters.

Similarly, there are 2 possibilities for nickels (0 or 1) and 3 possibilities for pennies (0, 1, or 2). So in all there are 4*3*2*3 = 72 possible combinations of coins.

However, some of these add up to the same amounts of money. For every combination that includes at least one quarter, but does *not* include the two dimes or the nickel, you could take away a quarter and add the dimes and nickel, and have the same amount of money. So, how many of these combinations are there?

There are three possibilities for the number of quarters: 1,2, or 3. There is one possibility for the number of dimes, 0, and one possibility for the number of nickels, 0. There are 3 possibilities for the number of pennies, 0, 1, or 2. So the number of combinations with at least one quarter, no dimes, and no nickels is 3*1*1*3 = 9.

So our original 72 combinations includes 9 combinations that are the same amount of money as another combination; that leaves 72-9=63 combinations that are all different amounts of money.

Finally, one of the possible combinations in our 63 is to have zero of each coin. The rules of the problem don't allow this, because we must use "some or all" of the coins, not none. So that leaves 62 valid combinations.