arrangement in a circle with a condition How many ways can we arrange 4 managers and 3 employees around a round table so that no 3 of the managers sit together...meaning 2 can sit together.
I know it involves the PIE principle and that if I do this:
total! - (arrangement in which all four managers are togther) - (arrangement in which three are together)...I will be over subtracting...I just don't get how to work around that...help will be highly appreciated
 A: Let us first count with numbered seats. Since solutions can be grouped into orbits of the group of $7$ rotational symmetries, each orbit of containing $7$ solutions each (since no solution can be invariant under all rotations, and $7$ being prime there is no nontrivial subgroup that could fix a solution), one can divide the result by$~7$ is we want to ignore this symmetry.
There are $7!$ ways to seat everybody without restrictions (I am assuming individual managers and employees can be distinguished from each other, as people usually can be). There are $\binom 43=4$ different triplets of managers one can form, for each of these we shall subtract off the placements where they sit together. First we order the triple in one of $3!=6$ ways, then we choose a seat for the first of them in $7$ ways; this fixes the places for the other two, after which $4$ people remain to be seated in $4!$ different ways. All in all we subtract off $\binom43\times3!\times7\times4!$ placements. Some placements however will be subtracted off more than once, because all $4$ managers are seated together. If this happens, there are exactly two groups of $3$ managers that sit together, so the placement was counted exactly twice, and to compensate we add back such configurations. By similar counting there are $4!\times7\times3!$ placements where all managers sit together. As there are factors $7$ throughout and we are going to divide by that, I'll keep those factors apart; the amount of numbered placements is
$$
  7\times6!-\binom43\times3!\times7\times4!+4!\times7\times3!
 =7\times(6!-4!4!+4!3!)=7\times288
$$
and not distinguishing placements that are rotationally symmetric there are $288$ placements.
As an afterthought, one can see that there is not only a common factor $7$ but an additional factor $4!\times3!$ that divides everything, and this can be understood: in any placement one can permute the managers and the employees among each other. So it would have been more efficient to count the subsets of $4$ seats that can validly be assigned to managers, and then (apart from dividing by $7$) multiply by $4!\times3!$ afterwards. There are $\binom74=35$ subsets of $4$ in all. The invalid subsets can be generated by choosing a first seat for a triplet and one of the remaining four seats, in $7\times4=28$ ways. The cyclically consecutive sets of four seats need to be added back, a total of $7$. So $35-28+7=14$ valid subsets; one has $14\div 7\times4!\times3!=288$.
A: Up to rotation and mirror reflection there are only two admissible ways to seat the employees: (i) on chairs $0$ and $\pm2$, and (ii) on chairs $0$ and $\pm3$. To get the total number of essentially different seatings of the seven individuals choose one of the employees as "middle employee" and put him on chair $0$, then choose one of (i) or (ii) and seat the remaining two employees accordingly (the choice involved here does not count), and finally seat the four managers arbitrarily in the remaining empty seats. This makes for a total of $3\cdot 2\cdot 4!=144$ essentially different seatings.
A: Take it this way. You can see that there is 0 way in which no two managers are seated together.Because for all the managers to sit separate there should be at-least 4 employees. 
Now, the no. of ways in which two managers can sit together includes the no. of ways of selecting any two managers from the four and the no. of permutations around the round table. 
So, the final answer would be 144.
A: Since this question has generated 16, 72, 144 and 288 as potential answers, I did a brute force check in Mathematica. This helps to straigthen my thinking.
Marc's 288 is correct. Alternatively, the answer "2" would be also be correct.
The two answers are eeMMeMM and eMeMeMM for faceless Managers and employees, since the question was open to this strict interpretation. 
Counting distinct individuals:
since there are less 'e' then 'M', the circular arrangement can be turned into a linear one by starting the (linear) count with e1.(**)
The first seating arrangement then allows e1e2MMe3MM and e1e3MMe2MM or 2*(4!)=48 arrangements.
The second seating allows e1MMe2Me3 and e1MMe3Me2 , again 48 arrangements.
Lastly, we must choose one of the 3 employees to play the role of e1,
giving 3*2*2*(4!) or 3*(48+48) = 288 arrangements.
(**) the 2 distinct seatings become 6 if we only require some employee as starting point along the table:
eeMMeMM , eMeMeMM , eMeMMeM , eMMeeMM , eMMeMeM and eMMeMMe.
