How many ways of putting 4 distinct pens in 3 identical boxes are there?

Question

I have 4 distinct pens and 3 identical boxes.I want to find the number of ways of putting those pens into these boxes,considering that there are no limitations(a box can contain no pens or all pens)

Answer 1

Since this has had a number of false (or at least misleading) answers posted, let me try to clarify.

The difficulty lies in trying to get the assumptions down. The problem tells us that the pens are distinct but the boxes are not. Thus "permuting the boxes" does not change any given solution. To be precise, putting all the pens in one box gives a single solution. It makes no difference which box we choose, as they are all the same.

To stress: the assumptions are critical here. If you change them, the answer will be very different. Here we are assuming that the pens are distinct but the boxes are identical.

If we ignore, for the moment, the fact that the pens are distinct, we see that there are exactly $4$ possible patterns. These correspond to those partitions of $4$ of length at most $3$. They are $$4\quad 3+1\quad 2+2\quad 2+1+1$$

Now we have to deal with the fact that the pens are distinct. How many ways can we populate each pattern? We'll do them one at a time.

Pattern $\underline 4$: There's only one way to put all the pens in one box, so $\boxed 1$.

Pattern $\underline {3+1}$: This allocation is entirely determined if you specify which pen is off on its own. Thus $\boxed 4$.

Pattern $\underline {2+2}$: Here, we get a wrinkle. There are $\binom 42=6$ ways to specify a pair of pens but there is a symmetry. The allocation $(AB,CD,0)$ is the same as the allocation $(CD,AB,0)$. Thus we need to divide by $2$, so $\boxed 3$.

Note: another way to see that the answer is $3$ in that case is to remark that the allocation is entirely determined by specifying which pen is paired with pen $A$.

Pattern $\underline {2+1+1}$: Now it is enough to just pick the pair which comprises the $2$, so $\boxed 6$.

The answer, then, is $$1+4+3+6=\boxed {14}$$

As the list is so short, let's just write them all out. Letting the pens be $\{A,B,C,D\}$ we have $$(ABCD,0,0)$$ $$ (ABC,D,0)\quad (ABD,C,0)\quad (ACD,B,0)\quad (BCD,A,0)$$ $$(AB,CD,0)\quad (AC,BD,0)\quad (AD,BC,0)$$ $$(AB,C,D)\quad (AC,B,D)\quad (AD,B,C)\quad (BD,A,C)\quad (BC,A,D)\quad (CD,A,B)$$

Answer 2

There are $81$ ways to do it if the boxes are distinguishable. Now, $3$ of those $81$ have "4 pens in one box", which corresponds to just one "identical-box" arrangement. Each of the other identical-box arrangements get counted $3!=6$ (not just $3$) times when we distinguish the boxes, because the pens are distinguishable and the only way to have multiple boxes with identical contents is to have two empty boxes. Thus the $81-3=78$ non-4-pens-in-one-box arrangements comprise $78/6 = 13$ identical-box arrangements. So there are $1+13=14$ such arrangements altogether.