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. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

Answers: 6, 24, 120, 360, 720

Correct answer is D, 360.

There is a great way to solve it in the explanations:

Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.

The correct answer is D.

But during the exam I didn´t even think about such a shorcut could be possible. I started to draw diagrams with the possibilities but because it was taking me too long I decided to pick an answer choice and keep going.

What I did during the exam was this:

If we named Frankie=F, Joey=J and the others 1, 2, 3, 4


J F _ _ _ _ That could be

JF1234
JF1243
JF1342
JF1324
JF1432
JF1423

We have 6 options with J F 1 _ _ _ in first positions, so we´ll have 6 more with J F 2 _ _ _, and so on 6x4=24

So once we correctly settles JF we have 24 possibilities with the other 4. Now we have to find all the possibilities to settle JF.

J F _ _ _ _
J _ F _ _ _
J _ _ F _ _
J _ _ _ F _
J _ _ _ _ F
_ J F _ _ _
_ J _ F _ _
.................

So on we get 15 possibilities

And 15X24 we correctly gets 360. Number D. I was doing right but I quit because it was taking too long, I chose 120 and Obviously I got it wrong.

Is it possible to do this problem in and statistical way so it is not as simple as the shorcut and not as long as this way?

Thanks

Here is how I solved it: -

Total possible cases satifying the criteria of Frankie behind Joe: -


1st Case: J _ _ _ _ _
2nd Case: _ J _ _ _ _
3rd Case: _ _ J _ _ _
4th Case: _ _ _ J _ _
5th Case: _ _ _ _ J _
6th Case: _ _ _ _ _ J

In all the above cases Frankie can take all the positions of '-'. In the grid above,I calculated the cases that satisifies the criteria Frankie behind Joe in all the possible cases: -

1st column 5! = 120 ways
2nd column, Frankie can't be ahead of Joe, so possible cases, 5! - 4! = 120 -24 = 96
3rd column, Frankie can't take be at the first two positions, so 5! - 2X4! = 120 -48 = 72
4th column, Frankie can't take the at the first three, so 5! - 3X4! = 120 - 72 = 48
5th column, Frankie can't take be at first 4, so 5! - 4X4! = 120-96 = 24
6th column, won't satisfy the criteria.

Hence total possible ways = 120+96+72+48+24 = 360

I know, it doesn't answers your original question, but still a different approach, and not very time consuming.

Hope it helps

GMAT 2007

Yeap, it does because it is shorter actually.

Thanks a lot.

Yes, nice explanation GMAT 2007. Luci, note that it is tough to come up with the best shortcuts the very first time you see a problem. Part of your study is to notice (after you've done it for the first time) that there is a shortcut (or read the explanation to learn it) and then think about how you will recognize the same shortcut on a similar problem in the future. (And the explanation can help to explain why the shortcut works.) Then, when you're taking the test, you recognize (as opposed to figure out from scratch) lots of shortcuts.

Very smart and elegant approach.

Consider the same logic in this case.

There are eight orators A,B,C,D,E,F.G & H. How mant ways can they speak at a function if C has speak before A and A has to speak before D and D has to speak before H .

1)40320
2)20160
3)1680
4)3360
5)6720

Shaji, are you posting that question because you want people to answer, or are you just offering another problem that could be approached using the above methods? FYI: if you are looking for a response from an instructor, you have to post the source of the problem. (Also, generally, it's a good idea to start a new thread if you want an instructor to answer, just to make sure it doesn't get lost in the mix.)

Stacey, I did post the question for people to apply the same or similar logic in a more intriguing version of a similar problem. The source of the problem is from Yahoo Groups, a forum for GMAT studies, but I did add a slight twist to make it more applicable for the logic used.

I thought Frank behind Joey meant - Frank is immediately behind Joey

JF_ _ _ _
_ JF_ _ _
_ _ JF _ _
_ _ _ JF _
_ _ _ _ JF

There 5 ways that JF can be together and 4! of arranging the others 5x4! = 120