# 10 people to be seated in 2 tables with 4 and 6 chairs respectively.

A person invites 10 guests at a dinner party and place them so that 4 are on one round table and 6 on the other round table. Find the number of arrangements.

My attempt

In first table with 4 seats, people can sit in $$1∗9∗8∗7$$ ways.

In the second table with 6 seats there are 6 people remaining who are yet to be seated. So those 6 guests can be seated in 1∗5∗4∗3∗2∗1 ways.

So Total arrangements = $$\frac{9!}{(6!)}∗5!$$

Where did that $$1$$ come from?

As it is a circular arrangement, the arrangement of the first person will not change anything , i.e, whatever seat he is placed at, he will have the same view. So the arrangement of the people matters from the second person. Hence, 1.

What am I doing wrong?

• It is wholly unclear what is meant by one "way" to seat the guests. Does it mean who's in exactly which chair? Who is to the right and to the left of whom? Who is next to whom? Who is at the 4 person table and who is at the 6 person table? These all give different answers. Aug 6 '19 at 19:31

• Pick which four people will sit at the smaller round table in $$\binom{10}{4}$$ ways.
• Now... let the youngest person who you selected sit at the table wherever they like, it matters not where. Then, after the youngest has chosen his seat, arrange the remaining people around the rest of the table in $$3!$$ ways.
• The youngest person from those not selected, let him sit at the larger table wherever he likes, it matters not where. Then, have the remaining people sit around the outside of that table in the remaining seats in $$5!$$ ways.

$$\binom{10}{4}\times 3!\times 5!$$

Your error is in neglecting to fully account for who the people sitting at the first table is. The "multiplication by 1" makes some sort of sense if the selection of the people has already occurred, but you jumped the gun with it, having used it before we know the full list of people sitting there yet.

Summarizing comments from below for easier access:

Another popular way of thinking of this type of problem is via a "division by symmetry" argument. A division by symmetry argument generally goes as follows: If we count the number of outcomes while allowing ourselves to overcount and we get an answer of $$n$$ including the overcounts and we learn that every outcome was overcounted the exact same number of times, having been counted $$k$$ times each, then the corrected count without overcounting would be $$n/k$$.

I tend to avoid these style arguments as a matter of principle, as they are a common source of headaches for people just starting, are challenging to phrase correctly, and require the use of division which would if used incorrectly take us out of the realm of the integers into the realm of the natural numbers. To a beginner, it is not clear at a glance why $$\dfrac{(kn)!}{(k!)^n}$$ should be an integer, but writing it as $$\binom{kn}{k,k,\dots,k}$$ it is much clearer that it will be an integer. That all being said, there are times where a division by symmetry argument is helpful and perhaps even clearer than the alternatives.

For this problem, let us starting from the small table and starting from north going clockwise select a person to sit in each seat, then once that table is full do the same for the larger table again starting from north. You will find there are $$(10\times 9\times 8\times 7) \times 6\times \cdots \times 2\times 1 = 10!$$ such arrangements.

Then, we recognize that each way of arranging the small table was overcounted by a factor of $$4$$ due to the different ways of rotating the table. Rather than $$10\times 9\times 8\times 7$$ it should have been $$10\times 9\times 8\times 7 / 4$$. The division by $$4$$ can be thought of as "forgetting which way was north" for that table. Note that the division is by $$4$$, not by $$10$$, because what we are dividing by was the number of times we had overcounted each occurrence. Similarly for the second table, we divide by $$6$$ because that was the number of different equivalent rotations of the table.

We get as a final result, the equivalent way of writing the final answer as:

$$\dfrac{10!}{4\times 6}$$

• @kaushik because there are ten people and we want to choose four of them to sit at the first table. Your answer effectively is where the youngest person always sits at the smaller table. Aug 6 '19 at 20:05
• @kaushik perhaps a "division by symmetry" arguement will make more sense to you. I prefer to avoid them out of principle. There are $10×9×8×7$ ways to pick and arrange the people for the small table if we cared who was sitting to the north. Then, we recognize there are four equivalent arrangements that should all have been counted as the same arrangement corresponding to the four rotations of the table, so we divide the result by four, not by ten, to account for the overcounting. Aug 6 '19 at 20:18
• Note, $10×9×8×7/10 = \binom{10}{4}×3!$. On the left, we place four of our ten people in a row around the circle starting from north and then "forget" which way north was. On the right, we pick four people simultaneously, then let youngest sit wherever, and then arrange the remaining people in the remaining seats. Both accomplish the same thing. Aug 6 '19 at 20:26
• I made a stupid typo in that last comment, which if you were following the convo should be apparent. It should have been 10×9×8×7/4, dividing by four, not by ten Aug 8 '19 at 10:36
• Eh... If the intention is good it usually doesn't matter much and if it gets really out of hand then mods will clean up. Having alerted me might have let me catch the typo in the window if time where edits were allowed. In any case, cheers Aug 8 '19 at 10:40

It will help if we start from scratch. The number of ways to seat $$n$$ people in a circle is $$(n-1)!$$, which may be seen by considering the viewpoint of an arbitrary person seated there. There are also $$\binom{10}4$$ ways to assign people to the smaller table, so there are $$\binom{10}43!5!=\frac{10!}{4\cdot6}$$ ways to arrange the people.

• I am sorry but I didn't understand the second part of your answer. Is it possible to elaborate a little? Aug 6 '19 at 19:42
• So what you mean is that, $\binom{10}4$ is for selecting 4 people out of 10 to sit at the table with 4 seats. But there are $3!$ ways to arrange them in that table. Now, 6 people are remaining and there are 6 seats. So they can be arranged in $6!$ ways. Is my understanding right? Aug 6 '19 at 19:48
• @Kaushik Once you fix the position of this "arbitrary person", the remaining $n-1$ lie in a line and may be permuted in $(n-1)!$ ways. As to assigning the people to tables, we are simply picking $4$ out of the $10$ to sit at the smaller table, and then the other $6$ go to the second table. Aug 6 '19 at 19:49