A Problem on circular permutation.

Question

We need to find the number of arrangements of 6 people at three identical round tables if each table must be occupied.

I tried to break it down into cases, like (1,1,4) , (1,2,3) , (2,2,2) as they are identical, the order won't matter in terms of number of people on a table. Also i am aware of the fact that number of ways of arranging $n$ people in a circle is $(n-1)!$. Also, i know if we need to arrange x objects in a line, where $ x_1 $ are of one kind, $x_2 $are of another kind and so , number of ways would be $= \frac{x!}{x_1!x_2!...x_k!}$ but i am not sure how to string these facts together for this problem.

Like for example of case (1,2,3) , first $\binom{6}{3}$ to select three people for first table , then $\binom{3}{2}$ to select two people for second table which automatically selects 1 person for first table. Then $\frac{\binom{6}{3} \cdot \binom{3}{2} \cdot 2}{z}$ but i am not able to figure out what that $z$ is going to be.

I would appreciate it if along with the answer you could also tell me method to handle these problems.

Answer 1

Looking first at the case $(1,2,3)$, you could call the table with $3$ people $A$, the table with $2$ people $B$, and the table with $1$ person $C$. The number of assignments of people to tables equals the number of arrangements of the word $AAABBC$ and is $\frac{6!}{3!\,2!\,1!}$ or $\binom{6}{3}\binom{3}{2}\binom{1}{1}$. I see both sorts of expressions in your attempted solution, except that your expression involving binomial coefficients has a factor $\frac{2}{z}$ that I don't understand.

At any rate, the next step is to arrange the people at each table, which, as you mention can be done in $2!\,1!\,0!$ ways. So the result is $$ \frac{6!}{3!\,2!\,1!}\cdot2!\,1!\,0!=\frac{6!}{3\cdot2\cdot1}. $$

For a case like $(1,1,4)$ you have to account for the interchangeability of the two tables with $1$ person, so you get $$ \frac{6!}{2!\cdot4!\,1!\,1!}\cdot3!\,0!\,0!=\frac{6!}{2!\cdot4\cdot1\cdot1}. $$

You can do the third case by the same principle. You should know that there is a systematic analysis of this problem: your problem is equivalent to finding the number of permutations of $6$ objects that have $3$ cycles in their cycle decomposition. This is the Stirling number of the first kind $\left[6 \atop 3\right]$.