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There are $40$ students who must be assigned to $4$ supervisors. And each supervisor must have at least $8$ students, then how many ways students can be assigned?

I tried a lot but not able to think about it, it is not a homework problem. I am taking combinatorics in fall, so I started studying on my own.

N. F. Taussig
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Randhawa
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    For problems like this it is important to specify details. Are the students distinguishable from one another? Are the supervisors? Suppose I have three students and two supervisors and require that each supervisor have at least one student. What is the answer then? – lulu May 21 '18 at 23:07
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    This is an unfortunately tedious problem. I see little way around this beyond finding all of the different ways of distributing the amount of students to each advisor and then for each such way of distributing the amounts count how many ways the students could be assigned in those amounts. For example, you could have them split up as $8,8,10,14$ where the first supervisor gets eight, the second gets eight, the third gets 10 etc... which the students could be assigned in that fashion in $\binom{40}{8,8,10,14}=\frac{40!}{8!8!10!14!}$ number of ways. – JMoravitz May 21 '18 at 23:08
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    @Daugmented you will overcount doing it like that. – JMoravitz May 21 '18 at 23:08
  • Yep, sorry, that true... – Daugmented May 21 '18 at 23:10
  • @JMoravitz , third supervisor can also be assigned with 8 students and 4th with 16. Can you explain it little more – Randhawa May 21 '18 at 23:15
  • Yes, and it could be $(8,9,11,12)$ and it could be $(11,9,8,12)$ and it could be $(16,8,8,8)$ etc... like I said... it is an incredibly tedious approach. The approach I describe is again, you go through and you list out all $\binom{11}{3}$ different ways to have the amounts of students distributed among the supervisors, and then for each of those ways find the number of ways of actually distributing the students in that fashion, and add all of these together. – JMoravitz May 21 '18 at 23:15
  • @lulu , yes they will be distinguishable (students and supervisors) – Randhawa May 21 '18 at 23:16
  • You can use generating functions to get a solution. However, this is more-or-less taking JMoravitz's approach in disguise - not sure how much faster you can compute it. – Jair Taylor May 21 '18 at 23:18
  • If everyone is distinguishable then it is a very tedious problem indeed. Maybe Inclusion-Exclusion is feasible, though I am inclined to doubt it. – lulu May 21 '18 at 23:19
  • Inclusion-exclusion would still require you to keep track of the number of ways in which one or more supervisors gets $0,1,2,3,4,5,6,$ or $7$ students. At a glance, it appears just as if not more tedious than doing it directly. – JMoravitz May 21 '18 at 23:20
  • @JMoravitz Oh, agreed. My guess is that there is no way to avoid some grueling listing of scenarios. – lulu May 21 '18 at 23:21
  • @JMoravitz , one of my friend told me its 15C4 * 11C4 * 7C4 . Is this correct? – Randhawa May 21 '18 at 23:25
  • Very definitely not. $\binom{15}{4}\cdot\binom{11}{4}\cdot\binom{7}{4}\approx 1.5\cdot 10^7$. This is far less than $\binom{40}{10,10,10,10}\approx 4.7\cdot 10^{21}$, the latter number being the number of ways in which the students can be distributed among the supervisors so that the supervisors get exactly ten students each, which is only a small subset of the possible ways to distribute the students satisfying that each supervisor gets at least eight students each. The correct final answer should be somewhere on the order of $10^{23}$ or $10^{24}$. – JMoravitz May 21 '18 at 23:28
  • @JMoravitz , then I think I need to wait until Fall to ask from professor – Randhawa May 21 '18 at 23:32
  • Whether this is a homework problem or not, is not the point or the issue. You state emphatically that it is not a homework problem. All good and well, but your question is written like you copied it from a text book. Where is your input? Your take on the problem. We object to problem statement questions (written as though reading an imperative from an assigned text exercise), – amWhy May 21 '18 at 23:37
  • Were you preparing a year in advance for your Fall 2018 combinatorics class when you posted this question 5 months ago? – amWhy May 21 '18 at 23:39
  • @amWhy , I am following few books to grasp content. this is not a exercise question otherwise I would have had solution. – Randhawa May 21 '18 at 23:40
  • I would cry if this question was straight copied from a textbook. It sounds like the OP might have seen a similarly worded problem and changed the wording slightly or tweaked the numbers in such a way that it had terrible consequences and the total number of arrangements exploded into obscene numbers. The question where each supervisor has at least one student is easy in comparison. The question where students are indistinguishable is also quite easy. – JMoravitz May 21 '18 at 23:41
  • What you asked is an exercise, Randhawa, and you showed absolutely no work or thinking on your part, in the post. You assigned us, in effect, a problem to solve for you. – amWhy May 21 '18 at 23:42
  • @amWhy , the question you are pointing is from Discrete Mathematics took last fall. which had two lectures on permutation and combination – Randhawa May 21 '18 at 23:42
  • @JMoravitz , It is not from book – Randhawa May 21 '18 at 23:45

1 Answers1

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You can take a generating function approach here. That is, you let $x$ be an indeterminate and find the $40!$ times the coefficient of $x^{40}$ in $$\left(\frac{x^8}{8!} + \frac{x^9}{9!} + \cdots + \frac{x^{16}}{16!}\right)^4.$$

The reason this works is that when you expand the product in all possible ways, factoring out a factorial of the degree, each product has a multinomial coefficient. For example, taking $$\frac{x^{11}}{11!} \frac{x^9}{9!} \frac{x^8}{8!} \frac{x^{12}}{12!} = \frac{40!}{11!\, 9!\, 8!\, 12!}\frac{x^{40}}{40!} $$ and multiplying by $40!$ gives the coefficient $$\frac{40!}{11!\, 9!\, 8!\, 12!}$$ which is the number of ways of choosing groups where the first supervisor has $8$ students, the second supervisor has $9$ students, etc. You only have to go up to degree $16$ in each factor because no group can have more than $16$ people.

Of course, expanding this out completely is equivalent to adding up all the possible multinomials corresponding to how many students each supervisor takes as JMoravitz mentions. But software packages like Sage are optimized to do this kind of convolution pretty quickly.

Edit: In Sage,

p = sum([x^n / factorial(n) for n in range(8, 17)])
q = (p^4).expand()
print q.coefficient(x,40) * factorial(40)

prints $435451605680654896510320$.

Jair Taylor
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  • This one looks correct. As I am satisfied with solution, So, accepting as an answer but I will add some opinions in Fall, if I had. – Randhawa May 21 '18 at 23:47