I have 3 quarters, 2 dimes, and 2 pennies. If I select at least one coin, how many different amounts of money can I make?

Question

I have 3 quarters, 2 dimes, and 2 pennies in my pocket. If I select at least one coin from my pocket, how many different amounts of money can I make?

I checked out some of the other coin questions. Still puzzled. We have 7 coins, 3 of one kind, two of another, and two of a third kind. Selecting AT LEAST ONE COIN means there are 7 cases to consider: one coin, two coins...up to 7 coins for the last selection. Multiple cases require adding the results from each case.

The first case is easy: selecting one coin gives 3 possibilities-a quarter, dime, or penny. The second case: selecting two coins. Using logic, there are 6 possibilities: 2-q, 2-d, 2-p, qd, qp, dp, which gives 6 different amounts. I'm not including duplicates. ie there could be a second pairing of 2 quarters, but that would have the same value as the first two quarter pairing.

Case 3: Selecting 3 coins. If I use 7C3 (7 choose 3) = 35, how do I account for duplicates? I could divide by 3!x2!x2! but that gives a decimal number answer, which doesn't make sense.

Not sure how to handle the rest of the problem.

Answer 1

This problem is particularly easy because you cannot make the value of one type of coin by summing other coins.

So there are $3$ quarters, $2$ dimes and $2$ pennies. There are four options for the quarter: $0$, $1$, $2$ or $3$... and likewise for the other coins.

$4 \cdot 3 \cdot 3 = 36$ possible values, where none-none-none is allowed. Subtract that single invalid choice to get $35$.

Answer 2

As the coins of the same value are interchangeable, think about how many of each value you draw. How many choices do you have for the number of pennies? of dimes? of quarters? Now multiply and subtract the case you don't draw any.

Answer 3

Think of it as a factorization, 4 possibilities on the exponent for the prime quarter, 3 possibilities for the exponent of dime, and 3 possible exponents on the value of penny. so the number in question in question has $4\cdot3\cdot3=36$ divisors, 35 aren't unitary.

This is just like: $$2^3\cdot3^2\cdot5^2=1800$$