Find the number of ways to arrange 8 students with restriction

Question

8 students are arranged in a row. How many ways to arrange them if 3 particular students must be separated?

Answer 1

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.

Answer 2

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.

Answer 3

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 $$ \binom{6}{3}\cdot 3! \cdot 5! = 14400 $$