Seven cars of seven different models are going to park in a

BTGmoderatorDC 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

OA B

Source: Magoosh

Since cars P and Q must be parked together, we can consider them as one. So, we have 6 cars instead of 7 cars. Six cars can be arranged in 6! = 720 ways. Since cars P and Q can be arranged as PQ or QP, the total number of ways = 2!*720 = 1440.

Out of these 1440 ways, in half the number of ways, Car S would be parked on the right of P and Q, thus, the required number of ways = 1440/2 = 720 ways

The correct answer: B

Hope this helps!

-Jay
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BTGmoderatorDC 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

OA B

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