Six mobsters have arrived at the theater for the premiere of

[tim]

If you want to get some general learning out of this problem, it's that there are many approaches to any given combinatorics problem. Your job is to read about and discover as many as you can, and start learning what solution techniques are available and what situations you should use them for. The more of these you do the better you will get at the process - AS LONG AS you are actively working to process this information rather than just focusing on getting the problem at hand correct..

[gaurav.gaba1]

Here is another approach

J _ _ _ _ _

For Frankie and others are five ppl, and we have 5 positions available, so 5!

_ J _ _ _ _
We have one of the 4 places available for Frankie, and 4 positions for other 4 ppl.

So, it would be 4*4! ways

_ _ J _ _ _

Similarly, 3 places for Frankie and 4 postions for other 4 ppl.

3*4!

_ _ _ J _ _

2*4!
_ _ _ _ J _

1*4!
_ _ _ _ _ J

This case not possible

Adding all, we get 5! + 10* 4! = 120 +240 =360


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

GMAT 2007
can u explain this one
2nd column, Frankie can't be ahead of Joe, so possible cases, 5! - 4! = 120 -24 = 96

[proximityinfotech3]

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.

________________________________________________________

Office Chairs | Chairs

[skprocks]

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

For this explannation I do not understand,why we need to subtract from 5! the number of possible arrangements of F.
Say when -J---- ,F can stand behind J in 4! ways,right.
Why do we say 5!-4! ways.Please help me understand.
And further on,there are 5!-2*4!,5!- 3*4! etc...
What am I missing?

Anyhow the question wants to know the total no. of ways in which F is behind J.
So,answer shall be 5!+4!+3!+2!+1!=153.

[mschwrtz]

"Anyhow the question wants to know the total no. of ways in which F is behind J.
So,answer shall be 5!+4!+3!+2!+1!=153."

Here's the trouble with that approach: It assumes that once Frankie is in the second (for instance), you need only assign 4 gangsters to the four spots 3-6. But there are in fact four gangsters (all except F and J) anyone of whom could be assigned to the first spot. AFTER that assignment, any of the remaining four could be assigned to any of spots 3-6. This gets very messy once J is in the third or later spots.

" I do not understand,why we need to subtract from 5! the number of possible arrangements of F."

This finesses the difficulty above. There are 5! arrangements with J in spot 2. Of those, 4! have F in spot 1. 5!-4! leaves the acceptable arrangements. And so on.

[abn5n]

Here's how I solved it; hope this helps:

I added up all the different combinations for Frankie to be behind Joey.

1st scenario: Joey is first, so there is one option for the first spot, and 5! options for the remaining spot:
1 * 5 * 4 * 3 * 2 * 1 = 120

2nd scenario: Joey is second, so there is one option for the 2nd spot, 4 options for the first spot (1st spot can't be joey or frankie, so must be on of the other 4) and 4! options for the remaining spots (out of 6 possible people, we've used up Joey already and 1 of the other guys for the first seat, so there are 4 guys remaining to arrange:
4 * 1 * 4 * 3 *2 * 1 = 96

3rd scenario: Joey is 3rd, so there is one option for the third spot, 4 options for the 1st spot (can't be Joey or Frankie), 3 options for the second spot (can't be J, F, or the guy in spot 1), and 3! for the remaining spots (can't be J or the two guys up front):
4*3*1*3*2*1 = 72

4th scenario: J is in 4th spot, so there is one option for that spot, first three spots are 4*3*2 (based on logic above), and there are two options for the 5th spot (can't be J or the three guys up front)
4*3*2*1*2*1= 48

5th scenario: J is 5th, so there is one option for that spot, F only has one place he can go, 6th, so there is one option for that spot, the other guys are permutated up front:
4*3*2*1*1*1 =24

6th scenario: there is no 6th scenario as J can't be last!

Add up all the permutations:
120+96+72+48+24 = 360

[tim]

so now the real question you want to ask yourself is whether that was the most efficient way to solve the problem.. :)

[elyssaduboys]

Can you use the "stuck" together method for this problem?

[jnelson0612]

Can you use the "stuck" together method for this problem?

No, because they do not have to be next to each other. One of them could be in position 1 and one of them could be in position 6.

[700+]

can u explain this one
2nd column, Frankie can't be ahead of Joe, so possible cases, 5! - 4! = 120 -24 = 96

Consider the slot method

2nd Case:

_ <- Frankie can't come over here. Only remaining 4 mobsters can come over here.
J <- We are considering Joe in the 2nd row. So 1.
_ <- Now including Frankie, there are 4 mobsters remaining. Any of the 4 mobsters can come here.
_ <- Remaining 3 mobsters can come over here
_ <- Remaining 2 mobsters can come over here
_ <- Remaining 1 mobsters can come over here

Total no of arrangements in which Joe is in the 2nd row & Frankie is behind Joe = 4 x 1 x 4 x 3 x 2 x 1 = 96

Hope this helps.

[jnelson0612]

can u explain this one
2nd column, Frankie can't be ahead of Joe, so possible cases, 5! - 4! = 120 -24 = 96

Consider the slot method

2nd Case:

_ <- Frankie can't come over here. Only remaining 4 mobsters can come over here.
J <- We are considering Joe in the 2nd row. So 1.
_ <- Now including Frankie, there are 4 mobsters remaining. Any of the 4 mobsters can come here.
_ <- Remaining 3 mobsters can come over here
_ <- Remaining 2 mobsters can come over here
_ <- Remaining 1 mobsters can come over here

Total no of arrangements in which Joe is in the 2nd row & Frankie is behind Joe = 4 x 1 x 4 x 3 x 2 x 1 = 96

Hope this helps.

Thanks!

[aps_asks]

Hi Ron/Stacey

Can you please suggest a simpler approach to solve such problems ?

[jnelson0612]

Hi Ron/Stacey

Can you please suggest a simpler approach to solve such problems ?

The simplest way to do this problem is simply to think:

How many ways can the mobsters arrange themselves in line? 6 choices for the first position, then 5 for the second, then 4 for the third, and so on. This is 6!, which is 720.

In what fraction of these arrangements will Joey be in front of Frankie? Well, they can never stand side by side; one is always in front of or behind the other. Hmmm . . . isn't it reasonable to think that in half of the arrangements Joey will be somewhere in front of Frankie and in half of the arrangements Frankie will be somewhere in front of Joey?

If you have any difficulty conceptualizing this, think of just the two men, Frankie and Joey. They will either stand FJ or JF. The other people are just filler around and/or between them. Half the time F will be in front; half the time J will be in front.

Thus, 1/2 of these outcomes, or 360, are acceptable. Let me know if you have further questions.

[aps_asks]

Thanks !

[tim]

:)