A man has 2 pennies, 3 nickels, 1 dime and 2 quarters. How many different sums of money can he make using one or more of these 8 coins?
well, start listing them:
p = 1
n = 5
d = 10
q = 25
pp = 2
pn = 6
...
ppnnndqq = 77
Now count them up, after eliminating duplicate amounts
To find the number of different sums of money the man can make using one or more of these 8 coins, we can use a systematic approach. Let's break it down step-by-step:
Step 1: Start with the highest value coin - the quarter.
The man has 2 quarters, so he can make 2 different sums by using 0 or 1 quarter.
Step 2: Move on to the next highest value coin - the dime.
The man has 1 dime, so he can make 2 different sums by using 0 or 1 dime.
Step 3: Next, consider the nickels.
The man has 3 nickels, so he can make 4 different sums by using 0, 1, 2, or 3 nickels.
Step 4: Finally, consider the pennies.
The man has 2 pennies, so he can make 3 different sums by using 0, 1, or 2 pennies.
Step 5: Add up the number of different sums from each step.
2 + 2 + 4 + 3 = 11
Therefore, the man can make a total of 11 different sums of money using one or more of these 8 coins.
To find the answer, we can try listing all the possible combinations of coins and counting them.
First, let's start by using only one coin. The man can choose any of the 8 coins, so there are 8 possible sums of money.
Next, let's consider combinations of 2 coins. To do this, we need to choose 2 coins from the 8 available. We can find the number of combinations using the formula for combinations:
C(n, r) = n! / (r! (n - r)!)
where n is the total number of coins and r is the number of coins to be chosen. Plugging in the values, we get:
C(8, 2) = 8! / (2! (8 - 2)!) = 28 combinations
So there are 28 possible sums of money using 2 coins.
Similarly, we can calculate combinations of 3, 4, 5, 6, 7, and 8 coins.
C(8, 3) = 8! / (3! (8 - 3)!) = 56 combinations
C(8, 4) = 8! / (4! (8 - 4)!) = 70 combinations
C(8, 5) = 8! / (5! (8 - 5)!) = 56 combinations
C(8, 6) = 8! / (6! (8 - 6)!) = 28 combinations
C(8, 7) = 8! / (7! (8 - 7)!) = 8 combinations
C(8, 8) = 8! / (8! (8 - 8)!) = 1 combination
Adding up all these combinations, we get:
8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 255
So, the man can make 255 different sums of money using one or more of these 8 coins.