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!
Counting - Six mobsters
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My point is "If we remove the restriction that Frankie must stand behind Joey,..."jessicmw wrote: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!
In other words, if we ignore the "rule" that Frankie must stand behind Joey (and let every stand where they want) then the mobsters can be arranged in 6! (720) different ways.
Once we've shown that, without the restriction, the mobsters can be arranged in 720 different ways, we can apply the fact that in half of those 720 arrangements, Frankie is standing behind Joey.
Cheers,
Brent
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
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
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6! ways of arranging themBrent@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
Joey is in front of Frankie in half of the arrangements.
So 6!/2 = 360
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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)
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)
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Nice work, arunspandaarunspanda wrote: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)
I love how many different solutions there are to this question.
Cheers,
Brent
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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
Arrangement 5: Permutation of JxxxxF as one unit =4*3*2 =24[4*3*2 or 24 ways to arrange JxxxxF]
Thanks.
Arun
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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.
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.
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?? 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.
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.
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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!!!
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!!!
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I think I understand what you're driving at. Here's a simpler demonstration.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!.
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!.
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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:ssuarezo wrote:Brent:Brent Hanneson wrote:I'm not 100% sure if your solution is the same as mine. I'll post mine here as well: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 ?
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
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.
___ ___ ___ ___ ___ ___
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!
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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
so ans :- 6!/2= 720/2 = 360
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