How many different combinations of pennies, nickels, dimes, quarters, and half dollars can a piggy bank contain if it has 20 coins in it? How many different combinations of pennies, nickels,
dimes, quarters, and half dollars can a piggy bank contain if it has 20 coins in it?
 A: If the order matters, we can use the product rule to show each of the $20$ coins can be chosen in $5$ ways, for a total of $5^{20}$ ways, assuming you have enough of each type of coins.
If the order does not matter (which in the case it probably does not) you have to use an $r$-combination. The formula is $C(n+r-1, r)$. Here $n = 5$ and $r = 20$. So we get $C(24,20)$.
A: The piggy contains $20$ coins, that is, $20$ spaces.
For the first space, there are $5$ possibilities: it can be a penny, a nickel, a dime, a quarter or a half dollar. For the second space we have the same possibilities because coins can be repeated. For 2 spaces we would have $5.5=5^2$ possibilities for combining coins so for $20$ spaces we would have $5^{20}$ possibilities.
A: Neither of the top two answers are correct. The correct answer is C( 24 choose 20)= 10626 different combinations. Think about the difference of distinct and indistinct objects????
A: $5^{20}$ only works when order in which the coins are chosen matters. Seems to me that this problem is only asking how many possible configurations of quantities for each coin there are in the piggy bank! For example, 4  pennies, 5 nickels, 10 dimes, 0 quarters, and 1 half dollar would be counted once! If you did $5^{20}$ you would count that configuration multiple times. 
The following excerpt explains how to count such situations...
From Discrete Mathematics and its Applications 7th edition Rosen:
"There are $\binom{n + r − 1}{r} = \binom{n + r − 1}{n − 1}$ $r$-combinations from a set with $n$ elements when repetition of elements is allowed."
"Proof: Each $r$-combination of a set with $n$ elements when repetition is allowed can be represented by a list of $n − 1$ bars and $r$ stars. The $n − 1$ bars are used to mark off $n$ different cells, with the $i$th cell containing a star for each time the ith element of the set occurs in the combination. For instance, a $6$-combination of a set with four elements is represented with three bars and six stars. Here
$∗∗ | ∗ | | ∗ ∗ ∗$
represents the combination containing exactly two of the first element, one of the second element, none of the third element, and three of the fourth element of the set. As we have seen, each different list containing $n − 1$ bars and $r$ stars corresponds to an
r-combination of the set with n elements, when repetition is allowed. The number of such lists is $\binom{n − 1 + r}{r}$, because each list corresponds to a choice of the $r$ positions to place the $r$ stars from the $n − 1 + r$ positions that contain $r$ stars and $n − 1$ bars. The number of such lists is also equal to $\binom{n − 1 + r}{n − 1}$, because each list corresponds to a choice of the $n − 1$ positions to place the $n − 1$ bars."
Thus, an $r=20$-combination from a set of size $n=5$ should be equal to $\binom{24}{20}$.
A: Short & Simple Explanation: 
This is the case of "indistinguishable objects into distinguishable boxes" (think about it)
In such a case 'n' is number of boxes (one for each type of coin), 'r' is number of coins to be selected (20).
So, total number of combinations is C(n+r-1, n-1) = C(5+20-1, 5-1)
