A boy 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?

Answer 1

He can use the 2 pennies in 3 ways, that is,
not take it at all, use one of them or use both of them
He can use the 3 nickels in 4 ways, that is,
not take any of them, take 1, take 2, or take all 3
etc.
So the number of sums you can get is 3*4*2*3 = 72
BUT, that would include the case of not using any of the coins which of course
would give you a sum of zero. So 72-1 = 71
BUT we have to consider the case where a sum can be obtained in two different ways.
e.g. 10cents --- 2 nickels, 1 dime, we should count that sum only once -- subtract 1
20 cents: 2 dimes, 1 dime + 2 nickels, --- subtract 1
25 cents: 1 quarter, 2 dimes and a quarter, 1 dime and 3 nickels --- subtract 2
30 cents: 1 quarter + 1 nickel, 2 dimes + 2 nickels --- subtract 1
11 cents: 2 nickels + 1 penny, 1 dime + 1 penny -- .....
12 cents: .....
16 cents:
17 cents:
.....

I will let you take over

or
you could just list the number of possible sums
1: 1 penny
2: 2 pennies
3: noway
4: noway
5: 1 nickel
....

Answer 2

no your wrong