Q: Ways of picking a collection of 12 coins from piles of pennies, nickels, dimes, quarters, and half-dollars given conditions: I am trying to solve the following problems, but I feel like my answers are incorrect.  I am hoping to get some verification and help with correcting them if necessary.
How many ways are there to pick a collection of 12 coins from piles of pennies, nickels, dimes,  quarters, and half-dollars?  Base on the following condition:
a)  Assuming that each pile has at least 12 or more coins.
Following the formula $(n+r-1)C(n-1)$
We take 12 coins, add the 5 combinations, and subtract 1, which equals 16.  Then we take 5 combinations - 1, which equals 4.  Hence:
$${16 \choose 4}=1820$$
b)  Assuming that each pile has at least 12 or more coins and the pick must consist of at least one quarter coin or at least one dime.
$${16 \choose 4}- {12 \choose 10}=1820-66=1754$$
c)  There are only 10 coins in each pile.
Wouldn't this just be the same as a?
$${16 \choose 4}=1820$$

A: For $(a)$, you are correct.
For $(b)$, I think it's easier to count how many have NO quarters AND NO dimes. Basically, that's choosing from 3 piles now... so $\binom{14}{2}$. Those are our invalid selections, so the answer is $\binom{16}{4} - \binom{14}{2}$.
For $(c)$, it's definitely not the same. You can't pick 12 pennies, for instance. Or 11 dimes and a quarter. All the piles only have 10 coins! So you could just subtract off the number of selections that had 12 of one coin (5 combinations) and 11 of one coin (20 combinations). 
