Find the number of ways to arrange 8 students with restriction 8 students are arranged in a row. How many ways to arrange them if 3 particular students must be separated?
 A: Select one of the $5$ non-particular students who will stand to the left of the rightmost particular student, and one of the remaining $4$ non-particular students who will stand to the right of the leftmost particular student. Permute the remaining $6$ students in $6!$ ways and place the two preselected non-particular students as planned, for a total of $5\cdot4\cdot6!=14400$ arrangements.
A: Hint:Find the no. of ways 2 particular students will be always together and twice the no. of ways 3 particular students will be always together.Subtract the 2nd from 1st.Subtract the result from the no. of all possible permutations.
A: Use stars to represent the five non-particular students and bars to represent the three particular students.  Then a feasible seating arrangement involves choosing three positions from the six that are either between two stars or at the end.  For instance,
$$
\star | \star \star \star | \star | 
$$
The number of ways to do this is $\binom{6}{3}$.  
Now keeping in mind the fact that the stars and bars are individual, distinguishable students, we can permute the stars and bars separately.  Therefore the number of arrangements is
$$
   \binom{6}{3}\cdot 3! \cdot 5! = 14400
$$
