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As in the title, I would like to distribute 10 distinct envelopes into 3 mailboxes. Let us also assume that some mailboxes can remain empty.

If they were identical envelopes, I know that we can use the stars and bars method, which would let us arrive at the answer of $\binom {12} {2}$.

However, here is where a friend and I started an extended discussion as we tweaked the question slightly.

What if, the envelopes were distinct and not identical?

My logic

Similar to the stars and bars method, 3 mailboxes is equivalent to 2 bars, so the question is equivalent to arranging 12 objects (10 distinct, 2 identical) in a line. Thus, my answer is $\frac {12!} {2!}$.

My friend's logic

My friend says that each envelope has 3 mailboxes to go into and since we have 10 envelopes, so the answer should be $3^{10}$.

Clearly, both of our answers are very different and I would obviously like to know who is correct and more importantly, why. We are currently at a standstill because neither of us can find the flaw in the other's logic. Any intuitive explanations will be greatly appreciated!

RobPratt
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Ethan Mark
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    Both envelopes and mailboxes being distinct, your friend's logic is correct. – true blue anil Mar 03 '21 at 06:31
  • @trueblueanil Where is the flaw in my logic then? I mean, yes, I assumed that the mailboxes were identical, hence I divided by $2!$, but even if they should be distinct, then why is $12!$ incorrect? – Ethan Mark Mar 03 '21 at 06:32
  • Your logic is incorrect. When you say $12!$, do you realize you are permuting envelopes within a mailbox too? – Math Lover Mar 03 '21 at 06:36
  • @MathLover Oh golly. Indeed. But why does my logic work with 10 identical envelopes and 3 identical mailboxes then? As mentioned in the question, we can use the stars and bars method to solve this, which would give $\binom {12} {2}$ and is equivalent to $\frac {12!} {10! 2!}$ (my method). Going by what you say, aren't I still "permuting envelopes within a mailbox"? – Ethan Mark Mar 03 '21 at 06:40
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    Firstly, I can't conceive a situation in which $3$ mailboxes (not just boxes) can be treated as identical. And though his starting logic is correct, the formulation leaves the possibility of two mailboxes being empty, so the answer is not correct. – true blue anil Mar 03 '21 at 06:45
  • Please read https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) and notice the point around bins being distinguishable as @trueblueanil mentioned. – Math Lover Mar 03 '21 at 06:51
  • When envelopes are identical, what you are doing is pretty much stars and bars. – Math Lover Mar 03 '21 at 06:54
  • @MathLover Yes precisely. Which is my confusion. Because you mentioned that if, in the case where the envelopes are distinct, my logic is flawed because I am permuting the envelopes within a mailbox right? I understand that now. But if that is the case, then how come I am allowed to use my logic for identical envelopes? Aren't I still permuting envelopes within a mailbox? Also, I have read the Wiki page quite a few times now. Haha. – Ethan Mark Mar 03 '21 at 06:59
  • Also, for identical envelopes and identical boxes, you will need to use partitions of a number. – true blue anil Mar 03 '21 at 06:59
  • @EthanMark No you are not. That is why you are dividing by $10!$ as you are considering them identical. – Math Lover Mar 03 '21 at 07:02
  • @MathLover I see. Okay yes. So basically the stars and bars method cannot ever work for distributing distinct objects then? – Ethan Mark Mar 03 '21 at 07:03
  • Yes it cannot. Because two distributions can differ only by the numbers of objects in each box. – user Mar 03 '21 at 10:35
  • @user Alright. I get it completely now. Thank you! – Ethan Mark Mar 03 '21 at 12:07

1 Answers1

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Stars and bars works fine when you have to distribute 10 identical envelopes into 3 distinct mailboxes, but not when the envelopes are distinct. Your argument gives the right answer ($12!/2!$) to the following question:

How many ways are there to arrange 10 distinct envelopes and two identical bars in a line?

If you regard the two bars as the dividers between three mailboxes, then under your approach the envelopes to the left of the first bar are sent to the first mailbox, the envelopes between the first and second bar are sent to the second mailbox, and the envelopes to the right of the second bar are sent to the third mailbox. The flaw in your approach is that you are overcounting what are equivalent assignments: once you've placed the bars, rearranging the items between the bars will lead to the same distribution of envelopes into the three mailboxes.

For grins, let's modify your reasoning to arrive at the correct answer ($3^{10}$):

For each of the possible ways to place the two identical bars, you are currently counting $10!$ permutations. This leads to a total of ${12\choose 2}10!$, which equals $12!/2!$. To repair this, replace $10!$ with a number that recognizes the overcounting going on. Specifically, suppose you place the bars at positions $a<b$. Then there are $a-1$, $b-1-a$, and $12-b$ envelopes going into the respective mailboxes. To fix the overcount, divide $10!$ by the number of ways to permute the envelopes between the bars. This leads to the total $$ \sum_{1\le a<b\le 12}\frac{10!}{(a-1)!(b-1-a)!(12-b)!}, $$ which equals $3^{10}$, by tedious calculation, or by the following

Claim: $$\sum_{1\le a<b\le N+2}\frac{N!}{(a-1)!(b-1-a)!(N+2-b)!}=3^N.$$ Proof: Write the sum in the form $$\sum_{a=1}^{N+1}\frac {N!}{(a-1)!}\sum_{b=a+1}^{N+2}\frac1{(b-1-a)!(N+2-b)!} \stackrel{(*)}=\sum_{a=1}^{N+1}\frac {N!}{(a-1)!}\sum_{c=0}^{N+1-a}\frac1{c!(N+1-a-c)!} $$ where in ($*$) we apply the change of index $c:=b-a-1$. The inner sum on the RHS equals $2^{N+1-a}/(N+1-a)!$, by the identity $${k\choose 0}+{k\choose 1}+\cdots+{k\choose k}=2^k,$$ which can be proved using the binomial theorem. To finish the proof, use the binomial theorem again to find the identity $$ 3^N=(1+2)^N=\sum_{i=0}^N{N\choose i}1^i2^{N-i}=\sum_{a=1}^{N+1}{N\choose a-1}2^{N+1-a}.$$

grand_chat
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  • As a sadistic mathematics lover, I actually grinned, but I will definitely stick to the simple $3^N$ :P and thank you so much for taking the time to write this! I will delete my own "answer" and give you the credit instead haha. Also, you mentioned that stars and bars work when I want to distribute 10 identical envelopes into 3 distinct mailboxes. Are the mailboxes really distinct though? Because in such a case the number of ways would be $\binom {12} {2}$ which is equivalent to $\frac {12!} {10! 2!}$ and by dividing by $2!$, does that not imply that the bars (i.e. mailboxes) are identical? – Ethan Mark Mar 04 '21 at 04:04
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    The bars represent the boundaries between the mailboxes/bins/boxes that you are tossing your items into. Though the bars are indistinguishable placeholder symbols, the boxes that they correspond to are distinct because we interpret the pattern of stars and bars spatially, as boxes reading from left to right -- for example the pattern $*||$ is considered distinct from its "mirror image" pattern $||*$ because in the first pattern the leftmost box has three items, while in the second pattern it's the rightmost box that has three items. – grand_chat Mar 04 '21 at 05:46
  • I see. So apart from the mathematics, a bit of contextualisation is also needed. Right. Thank you once again! – Ethan Mark Mar 04 '21 at 06:00
  • You are very welcome! – grand_chat Mar 04 '21 at 06:01