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For example with 15 people at 3 different tables each seating 5 people, is the number of combinations of seating equal to: $$\binom{15}5\binom{10}5(4!)^3$$

Parcly Taxel
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  • Hint: how many groups of what size do we have for this? If you answer that I am confident that similar to your approach you can work out the details for each of these cases. – Jan Oct 18 '18 at 12:50
  • Where did $4!$ come from? More broadly: what arrangements are the same? Are the seats numbered, say? (that's sort of what your formula suggests). – lulu Oct 18 '18 at 12:52
  • @lulu that was the example problem, not the title's. – Parcly Taxel Oct 18 '18 at 12:53
  • @ParclyTaxel Well, I think it is half and half. I think the $5's$ come from the example problem, and the $4$ was left over from the original. – lulu Oct 18 '18 at 12:54
  • if there are 4 different tables and 18 people, how many ways can the people be seated to have at least 4 in each table? I do not know how to solve this quesiton. –  Oct 18 '18 at 12:54
  • i tried many different methods but do not know. –  Oct 18 '18 at 12:55
  • The hint proposed by @Jan is the right way to start. What possible patterns are there? You could have $(4,4,4,6)$, say....what other patterns are there? Then count the ways to populate each type. (there really aren't many possible patterns). – lulu Oct 18 '18 at 12:56
  • I need the solution please. I am stuck. –  Oct 18 '18 at 12:58
  • Please try to follow the hint you have been given. – lulu Oct 18 '18 at 13:01

3 Answers3

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There are many different ways to approach the problem, depending on which "universe" of arrangements you are considering: people distinguished/undistinguished, same for seats, same for tables, plus whether the tables are round or linear.

The wording of the problem might be interpreted as un-dinstinguished people, un-d. seats, distinguished tables of unspecified type. That is: how to put 18 balls in 4 distinguished boxes, with no less than 4 balls in each.
If that is the correct interpretation, then two balls are free for being allocated wherever: $4 \cdot 4 /2$ ways.

When this happens in a test at school or elsewhere, it is advisable to accompany the answer with the "interpretation".

--- Addendum ---

There is some confusion going around here.

Let repeat @lulu's request and mine that when speaking of "ways" it is fundamental to specify which is the universe of equi-probable events being considered.

Let me try and clarify with an example. In how many ways can you flip a pair of fair coins ?

a) we consider only the cases "equal outcome" (HH or TT) and "different"(HT), because the coins are not labelled and we cannot distinguish (HT) from (TH). They are equi-probable, on the basis physical considerations.

b) three cases (HH), (TT), (HT): the coins are again not distinguishable; however three results are not equi-probable.

c) the coins are labelled and we distinguish the four cases (HH),(HT),(TT) and (TH).

Then are the "ways" 2, 3 or 4 ?

Note that, when we come to compute the probability as No. of favourable results / Tot. number of results, and provided the results are weighted appropriately, for the probability of having "equals" we get $1/2$ in every case.

Now, for the problem posed by OP, there so many possible assumptions that can be made, that an answer in terms of "ways" cannot be given unless carefully specifying the "universe".

G Cab
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  • would the solution be:18C4⋅14C4⋅10C4⋅6C6 multiply by3!⋅3!⋅3!⋅5! +18C4⋅14C4⋅10C5⋅5C5 multiply by 3!⋅3!⋅4!⋅4!= solution???? –  Oct 18 '18 at 14:50
  • So, to start 18C4 is the number of ways to accomodate $4$ distinguishable people out of $18$ into $4$ un-distinguishable tables: is this (part) of the assumptions you take as reference ?? – G Cab Oct 18 '18 at 15:18
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Using the fact that:

n people can be seated on a round table in (n-1)! ways.
Proof:
All n seats on a table are alike. Select any person to be seated first. He has only 1 choice of seats (there is only 1 way, whichever seat he chooses).
Now, symmetry is broken. So, remaining (n-1) seats can be considered as being in a line, with respect to the seat which was first occupied.
So, arrange remaining (n-1) people in a line in (n-1)! ways.

There are 2 possible arrangements: $$4,4,4,6$$ and $$4,4,5,5$$
For, $4,4,4,6$: divide 18 people in these groups:
$$^{18}C_4\cdot^{14}C_4\cdot^{10}C_4\cdot^{6}C_6$$ seat them in: $$3!\cdot3!\cdot3!\cdot5!$$
But, one table has 6 people while, other 3 have 4 each, so, select that one table in $^4C_1$ ways.

For, $4,4,5,5$: divide 18 people in these groups:
$$^{18}C_4\cdot^{14}C_4\cdot^{10}C_5\cdot^{5}C_5$$ seat them in: $$3!\cdot3!\cdot4!\cdot4!$$
2 tables have 4 while other 2 tables have 5 people. Select 2 out of 4 tables in $^4C_2$ ways. $$(^{18}C_4\cdot^{14}C_4\cdot^{10}C_4\cdot^{6}C_6)(3!\cdot3!\cdot3!\cdot5!)(^4C_1) +(^{18}C_4\cdot^{14}C_4\cdot^{10}C_5\cdot^{5}C_5)(3!\cdot3!\cdot4!\cdot4!)(^4C_2)$$ is the final solution.

pooja somani
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Given there are $ 4 $ different tables and $18$ people to be seated with at least $4$ in each table.

Let us number the tables $ 1,2,3,4$ as all tables are different.

if we put at least $ 4 $ in each table we have $ 2 $ extra persons.

Here we have only 2 possible configurations:-

  • $(6,4,4,4)$
  • $(4,4,5,5)$

For the first one:- $${^{18}C_6\times^{12}C_4\times^{8}C_4\times^{4}C_4\times{3!}^3\times{5!}\times{4}}$$

Here there are 4 different configurations of the selecting the table, ${\frac{4!}{3!}}$=$4$.

For the second case , $4,4,5,5$:- $${^{18}C_4\times^{14}C_4\times^{10}C_5\times^{5}C_4\times{3!}^3\times{4!}\times{6}}$$

Here there are 6 different configurations of the selecting the table, ${\frac{4!}{2!\times2!}}$=$6$.

The result is the sum of the above two cases.

sonorous
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  • would the solution be:18C4⋅14C4⋅10C4⋅6C6 multiply by3!⋅3!⋅3!⋅5! +18C4⋅14C4⋅10C5⋅5C5 multiply by 3!⋅3!⋅4!⋅4!= solution???? –  Oct 18 '18 at 14:50
  • No, there are not just 2 configurations, in the problem statement, it is not specified that you have to make all 18 seat at the same time. So there may be cases where not all 18 are selected which is dealt in the above explanation. – sonorous Oct 18 '18 at 14:59
  • so it would just be (3!⋅3!⋅3!⋅5!) + (3!⋅3!⋅4!⋅4!)? –  Oct 18 '18 at 15:03
  • but if it said all 18 needed to be seated, would it be the first response? –  Oct 18 '18 at 15:03
  • i am assuming that tables are identical, hence you have only two configurations (4,4,4,6) and (5,5,4,4) . – sonorous Oct 18 '18 at 15:40
  • you need to assume all 18 seats are occupied, unless mentioned otherwise. – pooja somani Oct 18 '18 at 15:41
  • and your tables seem to be distinguishable, you cant number them – pooja somani Oct 18 '18 at 15:44
  • @omega , in the above statement ,it is mentioned that tables are different, hence there would be 4 configurations related to (4,4,4,6), where 6 comes in every table, and for (5,5,4,4) , we have 6 combinations for different tables that must be taken into consideration. – sonorous Oct 18 '18 at 15:45
  • ok, if it is so. edited again :( – pooja somani Oct 18 '18 at 15:55