Seven cars of seven different models are going to park in

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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

[spoiler]OA=B[/spoiler]

Source: Magoosh

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by GMATGuruNY » Sat Jan 05, 2019 9:34 am
VJesus12 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
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.
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by Scott@TargetTestPrep » Fri Feb 01, 2019 5:24 pm
VJesus12 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

[spoiler]OA=B[/spoiler]

Source: Magoosh
If models P and Q must park next to each other, we can consider them as one unit. That is, there are 6 cars (instead of 7) and they can be arranged in 6! = 720 ways. However, since P and Q are considered one unit, they can be arranged as either PQ or QP. Therefore, there are 720 x 2 = 1440 ways to arrange these cars if there are no other restrictions.

However, we have a restriction that model S must be on the right of P and Q. Model S is to the right of P and Q in exactly half of the 1440 ways, and it is to the left of P and Q in the remaining half of the 1440 ways. Therefore, the number of ways S will be to the right of P and Q is ½ x 1440 = 720.

Answer: B

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