If you have 3 quarters, 5 dimes, and 6 nickles, how many distinct ways can you give someone $0.50 (Assuming that all denominations of coins are distinguishable from on another)?
I found 4 ways. Is that correct?
I think there are more.
2 quarters
5 dimes
6 nickles and 2 dimes
1 quarter, 2 dimes, 1 nickel
1 quarter, 1 dime, 2 nickels
I'll bet you can find more.
QNDD
QNNND
DDDDD
DDDDNN
DDDNNNN
DDNNNNNNNN
To find the number of distinct ways to give someone $0.50 using 3 quarters, 5 dimes, and 6 nickels, we can go through all possible combinations of coins.
One way to approach this is to start with the highest denomination coin (the quarter) and consider how many of them we can use. Since a quarter is worth 25 cents, we can use 0, 1, 2, or 3 quarters.
Let's go through the possibilities for each number of quarters:
1. 0 quarters: In this case, we have to make $0.50 using only dimes and nickels. Since a dime is worth 10 cents and a nickel is worth 5 cents, we can use 0 to 5 dimes, and the remaining amount will be made up of nickels. This gives us 6 possible combinations.
2. 1 quarter: In this case, we have a quarter and we need to make up the remaining $0.25 using only dimes and nickels. We can use 0 to 2 dimes, and the rest will be nickels. This gives us 3 possibilities.
3. 2 quarters: Similarly, if we have 2 quarters, we need to make up the remaining $0.50 using only dimes and nickels. We can use 0 or 1 dime, and the rest will be nickels. This gives us 2 possibilities.
4. 3 quarters: If we have 3 quarters, we don't need any additional coins since $0.75 is already greater than $0.50. So, there is only one possibility in this case.
Adding up the possibilities from each case, we have a total of 6 + 3 + 2 + 1 = 12 distinct ways to give someone $0.50 using the given coins.
Therefore, based on the explanation above, there are actually 12 distinct ways to give someone $0.50, not 4.