Question 1144454: How many ways can you make change of 94 cents using 50 cent coin,quarters,dimes,nickles and pennies? Can you show me the breakdown? I have 70 ways so far but need help. Thank you.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
It would be a miracle if you came up with the right number of ways by listing every combination in a random order. You need to make an organized list, so that you count each combination exactly once.
Your list is less than one-third of the way there; my count (quite possibly incorrect) was 219.
You will learn nothing by my solving the whole problem for you. I will show you how to make an organized list and work parts of the problem for you. Then you can tackle the whole problem if you want and see if you come up with the same number I did.
Let's develop a strategy before we start.
(1) All the coins except the pennies have values that are multiples of 5. With a total of 94 cents, that means the number of pennies has to be one of these numbers: 4, 9, 14, 19, ..., 74, 79, 84, 89, or 94. So break the problem into separate sub-problems for each of those numbers of pennies.
(2) For each case of a certain number of pennies, determine the possible combinations of 50-cent coins and quarters you can have. For each such combination, determine the amount remaining to be made using the dimes and nickels.
(3) When you get down to that point for each combination of pennies, 50-cent pieces, and quarters, it's easy to determine the number of ways to finish the 94 cents with dimes and/or nickels. Simply determine the number of choices you have for the number of dimes; any remaining amount will be made up with nickels.
Here is the process for one of the possible numbers of pennies.
29 pennies....
With 29 pennies, there are 65 cents left, so you can have a 50-cent piece and no quarters, with 15 cents left to be made using dimes and nickels. To make 15 cents using dimes and nickels, you have 2 choices for the number of dimes: 0 or 1. So there are 2 combinations using 29 pennies and 1 50-cent piece.
pennies 50-cent quarters remaining amount # of solutions
29 1 0 15 2
If you don't use a 50-cent piece with the 29 pennies, you can have either 0, 1, or 2 quarters.
With 2 quarters, there is again 15 cents remaining, with again 2 choices for the number of dimes to use. So 2 solutions using 29 pennies and 2 quarters.
With 1 quarter, there is now 40 cents remaining; you can have 0, 1, 2, 3, or 4 dimes, making 5 choices. So 5 solutions using 29 pennies and 1 quarter.
And with 0 quarters, there is 65 cents remaining, giving you 7 choices (0 to 6) for the number of dimes. So 7 solutions using 29 pennies and 0 quarters.
pennies 50-cent quarters remaining amount # of solutions
29 0 2 15 2
29 0 1 40 5
29 0 0 65 7
So there are 2+2+5+7 = 16 solutions using 29 pennies.
There will be larger numbers of solutions using fewer pennies, and fewer solutions using larger numbers of pennies.
For example, if the number of pennies is more than 69, you can't use 50-cent pieces or quarters, and all that is left is dimes and nickels.
So here is the complete analysis for numbers of pennies greater than 69.
pennies 50-cent quarters remaining amount # of solutions
74 0 0 20 3
79 0 0 15 2
84 0 0 10 2
89 0 0 5 1
94 0 0 0 1
Use the process demonstrated to finish the problem by analyzing the cases for the other possible numbers of pennies.
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