Problem Solving

Seven cars of seven different models are going to park in a row of seven side-by-side parking spots for an advertisement. Model P and Model Q must park next to each other, and Model S must be somewhere to the right of Models P & Q. How many possible configurations are there for the cars?

A. 600
B. 720
C. 1440
D. 4320
E. 4800

Answer B

P Q _ _ _ _ _
Ways = 1 * 1 * 5 * 4! = 120

_ P Q _ _ _ _
Ways = 1 * 1 * 4 * 4! = 96

_ _ P Q _ _ _
Ways = 1 * 1 * 3 * 4! = 72

_ _ _ P Q _ _
Ways = 1 * 1 * 2 * 4! = 48

_ _ _ _ P Q _
Ways = 1 * 1 * 1 * 4! = 24

Total Ways = 360

Since PQ can interchange their positions

So, 360*2 = 720

[spoiler]{B}[/spoiler]

sparkles3144 wrote:Seven cars of seven different models are going to park in a row of seven side-by-side parking spots for an advertisement. Model P and Model Q must park next to each other, and Model S must be somewhere to the right of Models P & Q. How many possible configurations are there for the cars?

A. 600
B. 720
C. 1440
D. 4320
E. 4800

Answer B

Let the 7 cars be A, B, C, D, P, Q and S.

Since P and Q must occupy adjacent positions, consider PQ a single element in the arrangement.
The number of ways to arrange the 6 elements A, B, C, D, PQ and S = 6! = 720.

In 1/2 of these arrangements, S will be to the LEFT of PQ.
In the remaining 1/2 of these arrangements, S will be to the RIGHT of PQ.
Thus, the number of arrangements in which S is to the right of PQ = (1/2)(720).

Since PQ can switch to QP -- doubling the total number of possible arrangements -- we multiply by 2:
(2)(1/2)(720) = 720.

The correct answer is B.