# Number of ways to distribute the coins?

There are 3 pots and 4 coins. All these coins are to distributed into these pots where any pot contain any number of coins.

In how many ways can all these coins be distributed if out of 4 coins 2 coins are identical and all pots are different?

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You really need to start putting (and showing) a bit more effort into your questions. You are merely copying statements, providing absolutely no evidence of any work on your part. The whole point is for you to learn how to solve these problems, and if you don't specify why you are unable to solve them (where you are stuck, what you have tried and why it doesn't seem to you to work, etc), you won't learn anything and you will soon exhaust any good will regular members of this forum extend to newcomers. – Arturo Magidin Apr 24 '11 at 11:58

Edit: I mistakenly computed with 4 pots rather than 3. Correct computation below.

• With 4 pots:

If both identical coins are together, then you select which pot the two coins go in (4 ways), and select where the other two, distinguishable coins go (4 ways for each), for a total of $4^3=64$ ways.

If the two identical coins are in different pots, then you select which two pots contain the coins ($\binom{4}{2}$ ways), then select where to put the other two, distinguishable coins (4 ways each), for a total of $\binom{4}{2}\times 16 = 96$ ways.

This gives a total of $64+96 = 160$ ways.

• With 3 pots:

As above: for both identical coins in the same pot, you have $3^3 = 27$ ways of distributing the coins.

For the identical coins in different pots, you have $\binom{3}{2}\times 3^2 = 27$ ways as well.

So the total is $27+27 = 54$ ways when there are three pots.

Added 2. Another way of thinking about it, which may be better for generalization purposes, is to think of the two identical coins as 3-sided dice that you are rolling, with the outcome telling you in which pots to put the coins. You want to count the total number of distinct rolls with two 3-sided dice to find out the number of ways of distributing those two coins. The other, distinguishable, coins each have 3 ways to be placed.

For two dice with three sides each, you are counting combinations with repetitions, so the formula is $$\binom{n+r-1}{n}$$ where $n$ is the number of dice, $r$ the number of faces in each die. So with $n=2$ and $r=3$, you get $\binom{3+2-1}{2} = \binom{4}{2} = 6$ ways of doing it. Then we have three ways for each of the other coins, giving you $6\times3\times 3 = 54$ total ways.

This method is easier to generalize to, say, four identical coins and three other different coins with 5 pots, without having to consider different cases. Such a problem would give you $\binom{4+5-1}{4} = \binom{8}{4}$ ways of placing the four identical coins, and $5^3$ for the remaining three different coins, for a total of $$\binom{8}{4}\times 5^3 = 8750$$ ways. Or if you have four coins and three pots, with 2 sets of two equal coin (say, two pennies and two dimes), then you would have $\binom{2+3-1}{2} = \binom{4}{2}=6$ ways for each set of two equal coins, for a total of $36$ ways. Simpler than considering the case where each set is together, each set is separated, etc.

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(4,0,0)->3 ways (3,1,0)-> 18 ways (2,2,0)-> 12 ways (2,1,1)-> 21 ways , so totel is 3+18+12+21=54.....What do you think....is this right logic...if coins are(A,A,B,C) – prem shekhar Apr 24 '11 at 9:58
you've computed with 4 pots, rather than three. – Thomas Andrews Apr 24 '11 at 11:09
@Thomas: Sigh. Thanks. – Arturo Magidin Apr 24 '11 at 11:47
@prem: I first did the wrong number of pots. Your method does work if you keep in mind the two separate coins, but you end up having to consider way more cases than I did. – Arturo Magidin Apr 24 '11 at 11:51
I am very glad.....to get such a simple solution.....Thanks.......My way of calculation was bit messy – prem shekhar Apr 24 '11 at 12:11