Object arrangement, 5 objects in 3 bins If five diferent objects arrange into three bins, what is the total numbers of ways to arrange the objects if:


*

*Empty bin allowed : There will be $3*3*3*3*3 = 243$ ways 

*Bin 1 has at least one object: 
(total ways - bin 1 empty ) = $243 - 2*2*2*2*2 = 211$ ways 

*Every bin has at least one object: .. I need help on this


What if the five objects are identical ?
 A: For the distinguishable balls, we use Inclusion/Exclusion. Call a distribution of balls bad if at least one bin is empty. 
As you calculated, there are $2^5$ bad arrangements where Bin 1 is empty, and the same number where Bin 2 is empty, and where Bin 3 is empty.
If we find the sum $2^5+2^5+2^5$, we will have double-counted the bad arrangements where Bin 1 and 2 are empty. We will also have double-counted the bad arrangements where Bin 1 and 3 are empty, also where Bin 2 and Bin 3 are empty. So the number of bad arrangements is $2^5+2^5+2^5-3$. 
If the balls are identical, we have an entirely different problem, which is solved in general by using the Stars and Bars method (please see Wikipedia). 
Remark: The numbers here are very small, so the techniques we used in the answer are perhaps overkill. For example, for indistinguishable balls, we can make an explicit list: $(1,1,3), (1,3,1), (3,1,1), (2,2,1), (2,1,2), (1,2,2)$.
For the distinguishable balls, we can divide into the $6$ cases mentioned in the above paragraph, and make a count of each, taking advantage of symmetry to shorten the calculation. This is more work than the Inclusion/Exclusion procedure that we used in the answer but has a more concrete feel. You are probably expected to use Inclusion/Exclusion, since Part 2 kind of sets it up. 
A: Two things you need to consider:


*

*How many combinations to pick the 3 fixed out of 5.

*How many combinations to place the 2 objects left.
The first one is given by "5 ncr 3" which is equal to 5*4*3*2*1/(3*2*1*2*1) = 10.
So (5 ncr 3)*3*3 = 90
