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?
 A: 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.
Added. Corrected:


*

*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.
