A problem from Principles and techniques in combinatorics (Chen Chuan-Chong, Koh Khee-Meng) requires to find the number of ways in which 6 people can be seated around 2 tables (there is at least 1 pearson for each table).
The solution given starts by dividing the possibilities into 3 cases: 5-1, 4-2, 3-3. There are ${6 \choose 5}$ ways to divide six people into groups of sizes 5 and 1 each; thus the number of ways in this case is:
${6 \choose 5}*4!*0!=144$ (since the number of ways to arrange n objects in a circular permutation is $(n-1)!$)
By similar thinking we find out that the number of ways in the case of 4-2 people is:
${6 \choose 4}*3!*1!=90$
(And here is my question) At this point I was expecting the number of ways in the last case to be:
${6 \choose 3}*2!*2!=80$
but the authors say it is instead to pay attention, because it is:
$1/2{6 \choose 3}*2!*2!=40$
and they apparently leave the reason to the reader for exercise, yet I can't figure out why. Could you give me a helping hand please?
It has not been specified explicitly, but it looks as if the tables are supposed to be indistinguishable. So A, B, C sitting at Table 1, and D, E, F sitting at Table 2, is the same as D, E, F at Table 1 and A, B, C at Table 2.
To put it another way, there are for example $\binom{6}{4}$ ways to divide $6$ people into two teams, one of which has $4$ people and the other of which has $2$ people. But the number of ways to divide $6$ people into two teams of $3$ each is $\frac{1}{2}\binom{6}{3}$. For there are $\binom{6}{3}$ ways to choose one team to wear blue uniforms, and the other to wear red uniforms. But if they strip off their uniforms, then the number of distinct groupings gets divided by $2$, since what used to be A, B, C wearing blue and the rest wearing red is now indistinguishable from what used to be A, B, C wearing red and the rest wearing blue.