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?
