Let 50 guests come to a wedding gala. How many ways are there for all guests to sit in 5 distinct tables of 10 people capacity each? The order in which the guests are seated does not matter.
I understand that this is a combinatorics problem as the guests' sitting order does not matter. On the other hand the tables are distinct (and thus their ordering does matter?) so I get that I need to use permutations too.
What I have thought so far is that for a given table there are:
$$\binom{50}{10}$$
But then again I have to consider the fact that tables are distinct too. What is the proper way to work around that problem after all?
${50 \choose 10}{40 \choose 10}{30 \choose 10}{20 \choose 10}$ - Choose the 10 occupants of the "red" table. Then the 10 of the "yellow", and so on.
If sitting order matters and the tables are round, you should multiply by $(9!)^5$.
If the tables are the same (it only matters who sits with whom, not at which table) you should divide by $5!$ (permitting the groups between the tables).