Combinations 600 level problem..Please help

[dddanny2006]

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?
a.6
b.24
c.120
d.360 -------Answer
e.720

I dont understand why we say that in ahl of the cases...Joey will be infront and in the otehr half he will be behind.

I dont get that logic.

Thanks

Dan

[Stuart@KaplanGMAT]

I dont understand why we say that in ahl of the cases...Joey will be infront and in the otehr half he will be behind.

I dont get that logic.

Thanks

Dan

Hi Dan!

You're right, this question is as much about logic as it is about math.

We can paraphrase the question (a great thing to do with long word problems) as:

If six people are standing in a line, how many ways can they be arranged so that person x is somewhere behind person y?

Now let's apply some logic: x will always be either somewhere ahead of y or somewhere behind y. Is there a reason why he would be ahead of y more often than behind? Or vice-versa? No! Permutations are always symmetrical - in other words, you can flip every possible arrangement around to create an identical, but backwards, arrangement.

So, it only makes sense that 1/2 the time x will be in front of y and 1/2 the time x will be behind y.

Therefore, to solve, we simply take the total number of possible arrangements and divide by 2.

Accordingly:

6!/2 = 6*5*4*3 = 30*12 = 360

choose (D)!


Stuart

[Patrick_GMATFix]

Stuart's answer is phenomenal!

My approach took about 2 minutes to execute:

If Joey is first, Frankie and the rest can stand however they want in spots 2-6 --> 5! ways.

If Joey is 2nd, the others can stand however they want (5!) but get rid of the portion of those arrangements that have Frankie 1st. This takes out 1/5 of possibilities, leaving 5! * 4/5 ways.

If Joey is 3rd, the others can stand however they want (5!) but get rid of the portion of those arrangements that have Frankie 1st or 2nd. This takes out 2/5 of possibilities, leaving 5! * 3/5 ways.

If Joey is 4th or 5th, according to the same logic, there will be 5! * 2/5 ways and 5! * 1/5 ways respectively for Frankie to stand behind Joey

Joey can't be 6th because Frankie would be unable to stand behind Joey.

Add up all possibilities, and you get :

5! + (4/5)5! + (3/5)5! + (2/5)5! + (1/5)5! = 5!(15/5) = 5! * 3 = 360