If there must be at least one person in each table, in how many ways can 6 people be seated in 3 tables? 
If there must be at least one person in each table, in how many ways can 6 people be seated in 3 tables?

I know there are three possible ways to split the set of people P into three distinct groups ($1+1+4,1+2+3,2+2+2$). Consider only the case $2+2+2$. At first I thought that the number of ways to choose the first pair of people to sit on a table is ${6\choose2}$ but for some reason it is $\frac{1}{3}\times {6\choose2}$. Can anyone explain why this is the case?

We consider the 3 cases mentioned above:
1) If they are split 4/1/1, there are $\dbinom{6}{4}$ ways to select the 4 people sitting together, and $3!$ ways to arrange them at their table, so this gives $\dbinom{6}{4}(3!)=15(6)=90$ possibilities.
2) If they are split 3/2/1, there are $\dbinom{6}{3}$ ways to select the 3 people sitting together, $2!$ ways to arrange them at their table, and $\dbinom{3}{2}$ ways to select the 2 people sitting together, so this gives $\hspace{.2 in}\dbinom{6}{3}\cdot2\cdot\dbinom{3}{2}=20(2)(3)=120$ possibilities.
3) If they are split 2/2/2, then there are $\displaystyle\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}$ or $\;5\cdot3\cdot1=15$ ways to pair up the 6 people.
Therefore there are a total of $90+120+15=$225 possible ways to do this.

This is s(6,3), a Stirling number of the first kind.
