Selection from linear and circular arrangements Let $P(n)$ denotes the number of ways of selecting $3$ people out of $n$ sitting in a row, if no two of them are consecutive and $Q(n)$ is the corresponding figure when they are in a circle. If $P(n) - Q(n) = 6$, then $n$ is equal to?
 A: We represent the position of a selected person with a green ball, and the position of a person who has not been selected with a blue ball.
Linear arrangements in which no two of three selected people are consecutive:  Place $n - 3$ blue balls in a row, leaving gaps between successive balls and at the ends of the row.  There are $n - 4$ gaps and two spaces at the ends of the row for a total of $n - 2$ spaces.  Select three of these $n - 2$ spaces in which to place a green ball.  Now number the balls from left to right.  The numbers on the three green balls are the positions of the three selected people.  Thus, there are 
$$P(n) = \binom{n - 2}{3}$$
ways to select three people from the row of $n$ people so that no two of the three selected people are in adjacent seats.  
Circular arrangements in which no two of three selected people are consecutive:  A circular arrangement corresponds to a linear arrangement in which the ends of the row are joined.  Thus, a permissible circular arrangement is a permissible linear arrangement in which the two people at the ends of the row have not been selected.  We showed above that there are $P(n) = \binom{n - 2}{3}$ permissible linear arrangements.
Next, we count the number of permissible linear arrangements in which people at both ends of the row are selected.  Again, place $n - 3$ blue balls in a row, leaving spaces between successive balls and at the ends of the row in which to place a ball.  Place a green ball at each end of the row.  That leaves $n - 4$ spaces between successive blue balls in which to place a green ball.  Select one of them.  Now, number the balls from left to right.  The numbers on the green balls represent the positions of the three selected people.  Hence, there are $\binom{n - 4}{1}$ permissible linear arrangements in which the two people at each end of the row are selected.  Therefore, the number of permissible circular arrangements is 
$$Q(n) = \binom{n - 2}{3} - \binom{n - 4}{1}$$
Hence, 
$$P(n) - Q(n) = \binom{n - 4}{1} = n - 4$$
Notice that the difference is the number of linear arrangements in which people at both ends of the row are selected. 
Setting $n - 4 = 6$ gives the desired result.  
