How many ways are there to distribute 32 indistinguishable objects into five distinct bins such that four bins have an odd number of objects and one bin has x objects such that x mod 3 = 2.

Initially, I realized that 4 of the boxes must have at least 1 object, and that one box must have at least 2 objects. To maintain the first four objects as odd, we must add pairs of objects to them and to maintain the last bin as producing x mod 3 = 2, we must add triplets of objects to them. But I am unsure where to go from there.

  • $\begingroup$ Well, don't forget that there are odd numbers $\equiv 2\pmod 3$, so one bin might discharge two constraints. $\endgroup$ – lulu Sep 23 at 21:11

First suppose the question means that bins 1,2,3,4 have an odd number and bin 5 has $x$ objects

Your idea regarding initially putting 1 object into each of bins 1 to 4 and 2 into bin 5 is very sensible. Notice also that 4 odd numbers plus $x$ makes 32 and so $x$ is even i.e. $x$ is 2 modulo 6.

You must now put 13 pairs of objects into the bins with either 0, 3, 6, 9 or 12 pairs into bin 5.

It will be easy for you to count the cases now. However, you need to decide precisely what the question is asking:-

@lulu raised the issue of whether one of the odd numbers can also be the $x$. Can it?

Also, remember that the bin with $x$ objects can be any of bins 1 to 5.

Continuing with the solution. You need to do 5 separate calculations:-

If $N$ pairs go into the $x$ -box, the number of solutions is the number of ways of distributing 13-$N$ identical pairs into 4 distinct bins.

You then multiply the sum of the five numbers by 5 to take account of the position of the $x$ box.

We have $\begin{pmatrix}16 \\ 3\end{pmatrix}+\begin{pmatrix}13 \\ 3\end{pmatrix}+\begin{pmatrix}10 \\ 3\end{pmatrix}+\begin{pmatrix}7 \\ 3\end{pmatrix}+\begin{pmatrix}4 \\ 3\end{pmatrix}$

which must then be multiplied by 5.

  • $\begingroup$ Any of the boxes is able to be the box full-filling the modulo constraint. The odd numbers can also be x $\endgroup$ – KengoL Sep 23 at 22:09
  • $\begingroup$ UPDATE: I reread the question, the only constraints is that four boxes are odd and the last box has the 2 mod 3 property. Any distribution of the objects which follows this is valid. $\endgroup$ – KengoL Sep 23 at 22:18
  • $\begingroup$ When you say "last box" do you mean one other than the odd boxes? $\endgroup$ – S. Dolan Sep 23 at 22:21
  • $\begingroup$ In that case the method given above works. Are you familiar with "stars and bars" type problems? $\endgroup$ – S. Dolan Sep 23 at 22:28
  • $\begingroup$ I am familiar. I am just alittle unclear about accounting for the number of pairs which are not 0,3,6,9,12 in box 5. Distributing pairs into each box without taking into the condition would just be (13 + 5 - 1)C(13), but I am still unclear as to how to take into account the # of pairs that cannot be added to the 5th box (of which there are 7) $\endgroup$ – KengoL Sep 23 at 22:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.