How many different combinations of pennies, nickels, dimes, quarters, and half dollars can a piggy bank contain if it has 20 coins in it?

Question

score 4 · Accepted Answer · edited Apr 25 '16 at 15:55

4

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 $C(24,20)$.

edited Apr 25 '16 at 15:55

Laurent Duval

6,452

answered Dec 01 '13 at 05:27

user99680

6,708

This is also assuming the coins are kept as an ordered set, which is not the case with piggy banks I've seen. – user127096 Apr 08 '14 at 15:48
@cheap effective diet pills: Well, yes, but this is a math problem, not a physics problem, and no specs on dimensions nor otherwise are given. – user99680 Apr 09 '14 at 00:31
1

But it does say combinations; not "permutations". – user127096 Apr 09 '14 at 00:33
@cheap effective diet pills: I'm not sure it is meant in the technical sense of the word--maybe the OP can clarify this-- but without a specific model/layout and dimensions for the piggy bank, I do not see how this issue can be addressed: which arrangements would be considered equal to each other in an "abstract piggy bank"? Can you suggest an unambiguous way of addressing this? – user99680 Apr 09 '14 at 01:54
Sure. An arrangement is determined by the number of coins of each denomination that it contains. – user127096 Apr 09 '14 at 01:58
Well, I think we should let the OP decide what s/he wants. S/he accepted my answer, so I assume that was the goal of the question. If s/he wants otherwise, or someone else wants to ask me a follow-up,I will change it accordingly. – user99680 Apr 09 '14 at 02:18
Besides, without knowing the specifics, we do not know whether the shape restricts the size of coins; this then becomes a whole "coin-packing" problem which I think is way out of the scope of the original question. Still, if the OP states otherwise, I will change my answer to fit the statement. – user99680 Apr 09 '14 at 02:29

score 2 · Answer 2 · answered Dec 01 '13 at 05:59

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.

score 1 · Answer 3 · answered Apr 09 '14 at 17:04

1

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????

answered Apr 09 '14 at 17:04

David

11

score 0 · Answer 4 · answered Nov 25 '14 at 01:13

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)

score 0 · Answer 5 · edited Apr 18 '14 at 01:44

$5^{20}$ 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 $\binom{24}{20}$.

Please remember to use MathJax when posting math: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — ml0105, Apr 18 '14 at 01:45

How many different combinations of pennies, nickels, dimes, quarters, and half dollars can a piggy bank contain if it has 20 coins in it?

5 Answers5