A photo is being taken of Central High School's nine-member

For the purposes of easiness i assign the letters to each type of people:
S - shorter guys
T - taller guys
M - misbehaving guy
C - coach

So our arrangements: for shorter guys is SSSSS = 5! is the total combination since the first guy could be put in 5 ways, second in 4, third in 3 etc. 5*4*3*2*1
In the back row we have two cases TMCTT and TTCMT in both cases CM are fixed because coach has to be in the center, so only other 3 T will be changing their places. There are 3 places for 3 T, so 3!=6.
Now we need to find the total number of combinations possible, we multiply 5! to 3! and we get 720. But considering that there are two possible ways for upper row we multiply the result by 2 and get 1440. Answer is D.

There are basically two groups.
A group of 5 shorter players, which can be arranged themselves in 5! ways in front row
A group of 4 taller players, of which one player who is mischievous would stand right next to the coach. Remaining 3 taller players can be arranged themselves in 3! ways.

The coach will be in the middle of the taller players and the mischievous player can stand at right or at left of the coach. So we have to consider both these scenarios.

Arrangement of 5 shorter players in front row AND Arrangement of 3 taller players in back row(when mischievous player is to the immediate left of the coach) OR Arrangement of 5 shorter players in front row AND Arrangement of 3 taller players in back row(when mischievous player is to the immediate right of the coach)

We know in Permutations, Combinations, and Probability, AND stands for Multiplication and OR stands for Addition

Arrangement of 5 shorter players in front row * Arrangement of 3 taller players in back row(while mischievous player is to the immediate left of the coach) + Arrangement of 5 shorter players in front row * Arrangement of 3 taller players in back row(while mischievous player is to the immediate right of the coach)

5! * 3! + 5! * 3! ----------> 2(5!3!) -------> 1440 ------Choice D

I can't seem to get the correct answer. I'm trying to solve this using the SLOT method.

Here's what I did:

Sol'n 1 (front row)
9*8*7*6*5/5!

Sol'n 2 (second row)
4*3*2/3!

Sol'n 3 (coach and misbehaved boy)
2 possible outcomes (to the left and right of the coach who is in the middle)

Multiplying the 3 sol'ns, I got 1,008 arrangements. How do I get the correct answer?

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.