In how many ways can 4 males and 4 females sit together in a round table with such that there are exactly 3 males who are seated next to each other? I have tried using this method:
first I took 4C3 to choose 3 men to be seated together, then I multiplied by 3! for the 3 men to switch among themselves. After which I grouped the 3 men into one unit. Counting the other 4 women and the remaining 1 men, I should have 6 units. Since this is a round table, I took (6-1)! to arrange this 6 units. I realised that this gives the possibility of including the case where 3 and 4 men could be seating next to each other since that remaining 1 men can be seating next to one of the 3 men in the rotation, so I subtracted the case where there are exactly 4 men who sits next to each other. However, I do not get the answer. Can anyone advise?
below is my working:
$4C3 \times 3! \times (6-1)!-(5-1)! \times 4! = 2304$. The correct answer is $1728$. I tried other methods and I got the answer but I don't understand why this does not work.
 A: Since the table is circular we can use the group of three men as a reference point, after which there are just three possibilities for the position of the fourth man – one, two or three women between him and the nearest man to his left (this can be counted manually). Thus the $(6-1)!-(5-1)!$ should be just $3$ instead, and all other factors are correct, leading to $1728$.
A: Here is another approach you can take.
Take the quietest man and seat him first and then seat women in $4!$ ways. Now the remaining $3$ men must be seated together between any two adjacent women - there are $3$ such spaces between women.
Finally arrange men in $4!$ ways.
So the answer is $ \displaystyle 4! \cdot 3 \cdot 4! = 1728$
A: For your subtracted off case (where four men are seated together), the fourth man could be left or right of the group of three men designated to be seated together. So you need to double your subtracted off term.
A: Your method is correct , but it is $\color{blue}{\text{deficient}}$ ! As you realize , it is asked that $\color{red}{\text{exactly}}$ one three consecutive men , but your solution gives us $\color{red}{\text{at least}}$ one three consecutive men.So , you need to use the formula of $\color{green}{\text{GENERALIZED VERSION OF P.I.E}}$
$\color{purple}{\text{Lemma:}}$ For each $1 \leq m \leq t$ , the number of elements in S that satify $\color{blue}{exactly}$ $m$ of the conditions $c_1, c_2,...,c_t$ is given by $$E_m=S_m -\binom{m+1}{1}S_{m+1}+\binom{m+2}{2}S_{m+2}-\binom{m+3}{3}S_{m+3}+.......+(-1)^{t-m}\binom{t}{t-m}S_{t}$$
In our question our case is three consecutive men sitting together and it is asked for exactly one case ,so $m=1$ , and $t=2$ because there are at most $2$ case where $3$ consecutive men.Then , $$E_1=S_1 -\binom{1+1}{1}S_{2}$$
$E_1= \binom{4}{3}3!\times(6-1)!$ ,as you calculated
$E_2= 4!\times 4!$ ,as you calculated
Now , by using the lemma : $$E_1=\binom{4}{3}3!\times(6-1)! - \color{red}{2}\times 4!\times 4!= \color{blue}{1728} $$
