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

OA: b

800_or_bust 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

So I got this question last night going through the practice set from Magoosh. The solution noted that this is probably more difficult than anything that you would encounter on the GMAT, and its categorized as Very Hard. I got it correct, but it was mostly just an educated guess. I wasn't sure how to deal with the restriction that P & Q had to be next to each other. So I knew I was going to have to guess. Intuitively, choices (D) and (E) seemed way too large. I was able to prove this by kind of placing a far upper bounds on the possible answer by noting that only 5 cars could be placed in the far right, and likewise, once one car was placed there only five cars could be placed in the first slot. The remaining slots I just placed the cars at random using 5!. Well, this is 5x5x5!, which equals 25*120 = 100*30 = 3000. And I knew the actual answer would be much lower than this rudimentary calculation, so that clearly eliminated (D) and (E), but left (A), (B) and (C).

I ended up selecting (B) because I've gotten pretty good at guessing on Magoosh's practice set even if I don't really know how to approach a problem. I figured with 1440 being two times 720, it looked like a likely trap answer, which it was (i.e. it looked like it was there to trick people who might forget to divide something in half). Plus, 600 being a multiple of 120 also looked like a distractor. I figured 5! = 120 likely played some sort of role in the answer, since all three were multiples of 5! but this didn't help eliminate any of them.

The solution they gave was to recognize that there are only 12 possible configurations in which we can place P & Q such that they are right next to each other (ignoring the requirement that S be to the right of them). Placing the remaining 5 cars at random yields 5! = 120 possible placements. So this gives 12 * 120 = 1440 possibilities. Finally, they noted that in exactly 1/2 of those S would be to the left of P & Q, so the final answer was one-half of this, or 720.

My question: are there any other approaches to a problem like this? This was a very tricky problem, so I'm open to other approaches that might be more intuitive to me. Is the official solution the only way to tackle this?

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.

Another approach (not as elegant, but easy to follow):

Suppose P and Q are the first two cars. We've got 2 arrangements for them, and 5! for the rest.

Suppose P and Q are cars #2 and #3. We've got 2 arrangements for them, FOUR cars (everything but S) that can be first, and 4! for the rest.

Suppose P and Q are cars #3 and #4. We've got 2 arrangements for them, 4*3 for cars #1 and #2, and 3! for the rest.

You can probably see the pattern now:

2*5! + 2*4*4! + 2*4*3*3! + 2*4*3*2*2! + 2*4*3*2*1

or 720.