Hi Brent,

You said, "If we remove the restriction that Frankie must stand behind Joey, then the mobsters can be arranged in 6! (720) different ways.
Now, of these 720 different arrangements, it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time."

How can Joey be standing behind Frankie if the question says "Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand"

Thanks!

Lets take a different angle to this problem.

Probability of Franky behind Joe is 1/2 ( he either stands ahead of him or behind him).

Total arrangements is 6! = 720.

Favorable arrangement is 6!*p(F_behind_J) = 720*1/2 = 360.

Use this technique to combine probability and combinations. Works easy

Brent@GMATPrepNow wrote:I recently came across this question, and I quite like it.

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
E. 720

6! ways of arranging them
Joey is in front of Frankie in half of the arrangements.
So 6!/2 = 360

There are five arrangements.
Arrangement1: Permutation of JF as one unit and remaining 4 = 5!=120
Arrangement2: Permutation of JxF as one unit and remaining 3 = 4! * 4=96 [4 ways to arrange JxF]
Arrangement3: Permutation of JxxF as one unit and remaining 2=3!*4*3=72[4*3 or 12 ways to arrange JxxF]
Arrangement 4: Permutation of JxxxF as one unit and remaining 1=2!*4*3*2 =48[4*3*2 or 24 ways to arrange JxxxF]
Arrangement 5: Permutation of JxxxxF as one unit and remaining 1=4*3*2 =24[4*3*2 or 24 ways to arrange JxxxxF]
Hence, total possible ways = 120+96+72+48+24=360

Correct option is (D)

Correction to my post:

Arrangement 5: Permutation of JxxxxF as one unit =4*3*2 =24[4*3*2 or 24 ways to arrange JxxxxF]

Thanks.

Arun

Though this is lengthy,for clear understanding i think this might help you.

1 2 3 4 5 6

Case 1: J cannot take first place because it won't satisfy the given condition.
Case 2: If J takes second place then F has only one place to take and remaining can arrange among themselves in 4! ways.

FJ(4!)=1*(4!)
Case 3:If J is in 3rd place then F has 2 places and remaining (4!)
2*4!
Case 4:If J is in 4th place then F has 3 places and remaining (4!)
3*4!
Case 5:If J is in 5th place then F has 4 places and remaining (4!)
4*4!
Case 6:If J is in 6th place then F has 5 places and remaining (4!)
5*4!

Finally total=4!(1+2+3+4+5)=360

I hope this helps you.

?? People seem to be sweating it out in the name of perseverance here! What if there's a third person who should stand behind F. Or say between J and F?

6 people => 6 places. order defined for 2 of them, so select 2 spaces i.e. 6C2. The rest...can be arranged in 4! ways.

6C2*4! = 360. Say there's a 3rd person, beiber and the both of them want to keep an eye on him (duh!) and keep him between the two of them. Now, the order is defined for 3 people...go for 3 spaces => 6C3

the rest...3! So total # of ways = 6C3*3! ie. 120.

So, in general we can say that if N number of items are to be placed with constraint of placing certain R items (R<N) in a fixed order, total number of combinations: P = N!/R!.

I would love to prove the theorem in this way:
First, place the R items in the correct order.
Now the 1st or remaining N-R items can be placed in any of the R+1 'gaps' around the R items. X0X0X0X0X (as in this diagram, 5 places around 4 '0's)
Next, 2nd of the remaining can be placed in R+2 places around the R+1 gaps and so on up to the last or Nth item.
So, there are total P=(R+1)*(R+2)*(R+3)*...*N no. of ways. Multiplying the numerator and denominator by R!, P can be expressed as: P=N!/R!.

Caution: 1. If R items are fixed, as in this problem said, it's cool enough, otherwise we may need to multiply another factor NCR to choose R items.
2. If items are placed in circular fashion P=(N-1)!/(R-1)!.
3. Maybe there are mistakes as this is first hand typed, experts please comment!!!

binit wrote:So, in general we can say that if N number of items are to be placed with constraint of placing certain R items (R<N) in a fixed order, total number of combinations: P = N!/R!.

I think I understand what you're driving at. Here's a simpler demonstration.

Suppose we want to arrange A, B, C, D, E, F, and G. E, F, and G can go anywhere, but A must come before B, B before C, and C before D.

All we do to find the arrangements here is determine where E, F, and G are placed. Once E, F, and G are placed, A, B, C, and D must be arranged in that order from left to right in the remaining spaces.

For instance, suppose we have these seven spaces:

_ _ _ _ _ _ _

If I place E, F, and G as follows

E _ _ G F _ _

then my arrangement MUST be

E A B G F C D

So in this case, I only need to determine the number of ways I could place E, F, and G into three spaces. Hence the number of arrangements is (7 choose 3) * 3 * 2 * 1, or 7!/(3!4!) * 3!, or 7! / 4!.

ssuarezo wrote:

Brent Hanneson wrote:

NikolayZ wrote:I think we need to find the number of ways that arrange Frankie and Joey sitting in one row.
xxx 1
xxx 2
Indeed number of possible arrangements is 720 (6!)
So, there are 360 ways in which F&J are sitting in different rows and 360 ways in which they are sitting in the same one.
Am i right ?

I'm not 100% sure if your solution is the same as mine. I'll post mine here as well:

If we remove the restriction that Frankie must stand behind Joey, then the mobsters can be arranged in 6! (720) different ways.
Now, of these 720 different arrangements, it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time.

So, the answer is 720/2 or 360.

Extension: If there were n mobsters, then the answer would be n!/2

Brent:

Sorry, I didn't get it .. why is it clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time? why half the time?

Thank you,
Silvia.

1. You know they Frank is in front of Joey half the time because 6! would give you all the possibblities to fill a lines like this:
___ ___ ___ ___ ___ ___

now you can either "look" from the left and see Frank ahead or "look" from the right and see Frank behind. Since factorial is not really about which direction you look in filling up these blanks, and there are "two views", it's proof that one will be ahead of the other half the time.

2. I thought of it like 6P2 - choosing how the two can be arranged in a line of 6 where order matters then dividing by two because one is ahead of the other half the time. Then the other 4 spots can be arranged 4! ways.

So, (6P2)/2 * 4!

Ans is 360

4!(5+4+3+2+1)

6! is 720 . Half of the times joe will be behind frankie and half of the times is vice versa.

so ans :- 6!/2= 720/2 = 360

D