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The Department of business administration in which $80$ students made the next experiment:

the students were divided into four groups of $19$. From each group we choose one worker, $3$ inspectors, $6$ Organizational Consultants, and other managers (9).

How many options for distributing the experiment?
A. when there is no limit.
B. When 'Jacob' and 'Michael' will not be on the same team, and Jacob must be elected to any group (never left out).

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  • $\begingroup$ "Jacob and Michael will not be in same team" does it mean they both cannot be among the inspectors/OCs/Mgrs from one group? Also, the managers (or OCs or inspectors or workers) from different groups are considered different teams or same team? $\endgroup$ – tpb261 Jun 7 '14 at 9:10
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    $\begingroup$ Only a Department of business administration could divide $80$ students into $4$ groups or $19$. $\endgroup$ – Marc van Leeuwen Jun 7 '14 at 14:21
  • $\begingroup$ please tell if i answered your question $\endgroup$ – RE60K Jun 7 '14 at 14:50
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First part: Given one group of 19 people, we have $$\binom{19}{1}\binom{19-1}{3}\binom{19-4}{6}$$ ways of forming the experiment. For four groups it becomes $$\left( \binom{19}{1}\binom{19-1}{3}\binom{19-4}{6} \right)^4$$Now the individual groups can be formed in $$\binom{80}{19}\binom{80-19}{19}\binom{80-38}{19}\binom{23}{19}\frac{1}{4!}$$ ways. So with no restrictions the experiment can be conducted in $$\binom{80}{19}\binom{61}{19}\binom{42}{19}\binom{23}{19}\left( \binom{19}{1}\binom{19-1}{3}\binom{19-4}{6} \right)^4\frac{1}{4!}$$ ways.

Second part: I'll try to find a simpler solution, but till then this can do, I guess.

We will find number of ways in which both can be in same team, and subtract that from total number of ways.

Since J has to be elected, let him be in the first group. The first group now needs 18 people from 79. The number of possible ways to form groups is $$S = \frac{1}{4!}\binom{79}{18}\binom{61}{19}\binom{42}{19}\binom{23}{19}$$ Total number of team formations = $$\frac{1}{4!}\binom{79}{18}\binom{61}{19}\binom{42}{19}\binom{23}{19}\left( \binom{19}{1}\binom{19-1}{3}\binom{19-4}{6} \right)^4$$

Number of ways in which J and M will be in one team: For this both are to be in same group, so number of groupings are reduced to: $$\frac{1}{4!}\binom{78}{17}\binom{61}{19}\binom{42}{19}\binom{23}{19}$$

Number of possible teams for that group, if both are in same team of:

  1. Inspectors = $\binom{19-2}{1}\binom{19-2-1}{3-2}\binom{19-3-1}{6}$
  2. OCs: $\binom{19-2}{1}\binom{19-2-1}{3}\binom{19-3-1-2}{6-2}$
  3. Managers: $\binom{19-2}{1}\binom{19-2-1}{3}\binom{19-3-1-2}{6}$

So total number of teams in which J and M will be in same team = $S_1 = \frac{1}{4!}\left(\binom{78}{17}\binom{61}{19}\binom{42}{19}\binom{23}{19}\right) \left(\binom{19}{1}\binom{19-1}{3}\binom{19-4}{6}\right)^3 \left( \binom{19-2}{1}\binom{19-2-1}{3-2}\binom{19-3-1}{6}+\binom{19-2}{1}\binom{19-2-1}{3}\binom{19-3-1-2}{6-2}+\binom{19-2}{1}\binom{19-2-1}{3}\binom{19-3-1-2}{6} \right)$

So number of ways in which J is always chosen into the teams and J and M will not be in same team = $S - S_1$

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  • $\begingroup$ thanks a lot. but I actually meant that both (J and M) would be on the same group (of 19). $\endgroup$ – user144555 Jun 7 '14 at 13:45
  • $\begingroup$ Can you explain your edits? And also why my way of forming groups is wrong? $\endgroup$ – tpb261 Jun 7 '14 at 17:46
  • $\begingroup$ @tpb261 i just centered everything to make it more readable. $\endgroup$ – RE60K Jun 8 '14 at 10:59
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Part A

$$A:\large\frac{^{80}C_{76}.76!}{4!}\frac1{(9!6!3!)^4}=80!.\frac1{(9!6!3!)^4.(4!)^2}$$ Suppose a class with four rows each of 19 benches.We label each bench with the (Group No.) and (Post). Selecting 4 out and arranging 76 students in these rows [$^{80}C_4.76!$]will fulfill the task.Also there will be multiple arrangements which will be same as if the goup 1 and group 2 interchange and then sit so we divide the ways by $4!$ because the groups are not unique like they have different names.Also we divide by $(9!6!3!)^4$ so to cancel internal permutation.

Part B

Similiarly: $$B:\large\frac{^{78}C_{74}.74!.^4C_1.^{19}C_2.2!}{4!}\frac1{(9!6!3!)^4}=78!.\frac{19.18}{(9!6!3!)^4.3!.4!}$$


Aliter:

Part A

Making four groups :$$\large\alpha=\frac{^{80}C_{19}.^{61}C_{19}.^{42}C_{19}.^{23}C_{19}}{4!}=80!\frac1{19!^4.4!^2}$$ For internal division:$$\large\beta=\left[^{19}C_{9}.^{10}C_{6}.^{4}C_{3}.^{1}C_{1}\right]^4=19!^4\frac1{(9!6!3!)^4}$$ Total ways: $$\large\alpha\beta=80!.\frac1{(9!6!3!)^4.(4!)^2}$$

Part B

Making four groups of (19,19,19,17): ['Jacob and Michael' in last group] $$\large\alpha^{'}=\frac{^{78}C_{19}.^{59}C_{19}.^{40}C_{19}.^{21}C_{17}}{3!}=78!\frac1{19!^3.17!.3!.4!}$$ For internal division: $$\large\beta^{'}=\left[^{19}C_{9}.^{10}C_{6}.^{4}C_{3}.^{1}C_{1}\right]^4=19!^4\frac1{(9!6!3!)^4}$$ Total ways: $$\large\alpha^{'}\beta^{'}=19.18.78!. \frac1{(9!6!3!)^4.3!.4!}$$

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  • $\begingroup$ Isn't $^{80}C_4.76!$ the number of permutations of 4 objects from a set of 80? It is not the same as what is being asked. $\endgroup$ – tpb261 Jun 7 '14 at 10:20
  • $\begingroup$ $^{80}C_4=^{80}C_{76}$ It is same if we remove 4 students and then arrange or if we select 76 students then arrange $\endgroup$ – RE60K Jun 7 '14 at 13:02
  • $\begingroup$ in this case, the two combinations: $G_1 = {S_1,S_2,...S_{19}}$ and $G_1 = {S_2,S_1,S_3.....S_{19}}$ would count as two different experiments, while in actuality they should be treated as only one, as the students in the group are not changing. $\endgroup$ – tpb261 Jun 7 '14 at 13:29

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