Find the Total Number of Arrangements for Six Mobsters in Line

  • Thread starter Thread starter Polymath89
  • Start date Start date
  • Tags Tags
    Permutation
AI Thread Summary
The discussion revolves around calculating the total arrangements of six mobsters in line, with the condition that Frankie must stand behind Joey. The initial calculation mistakenly limited Frankie’s possible positions based on Joey’s placement, leading to a total of 240 arrangements. However, it was clarified that if Joey is first, Frankie has five spots available behind him, not four, allowing for more arrangements. The correct total is 360 arrangements, as Frankie can occupy any position behind Joey, including directly behind him. This misunderstanding stemmed from a misinterpretation of the phrase "not necessarily behind him."
Polymath89
Messages
27
Reaction score
0
Question: 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?

This is how I tried to solve it. The number of arrangements depend on where Joey stand. So e.g. if Joey is the first in the row, then Frankie has to stand in one of the last four spots of the line so that you have 4! arrangements for those and one of the other 4 mobsters can take the second spot. So the total arrangements for this is 4*4!. Next if Joey is in the second spot, Frankie has to be in one of the last 3 spots, which gives a total of 3*4! arrangements. So in total I got 4!(4+3+2+1)=240 arrangements, however the answer to the question is supposed to be 360. Can anybody tell me what I did wrong?
 
Physics news on Phys.org
Polymath89 said:
Question: 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?

This is how I tried to solve it. The number of arrangements depend on where Joey stand. So e.g. if Joey is the first in the row, then Frankie has to stand in one of the last four spots of the line so that you have 4! arrangements for those and one of the other 4 mobsters can take the second spot. So the total arrangements for this is 4*4!. Next if Joey is in the second spot, Frankie has to be in one of the last 3 spots, which gives a total of 3*4! arrangements. So in total I got 4!(4+3+2+1)=240 arrangements, however the answer to the question is supposed to be 360. Can anybody tell me what I did wrong?

If Joey is first in line, then there are five spots behind him. Not four. Frankie CAN stand right behind Joey, he just doesn't have to.
 
Ah ok thanks a lot. I thought "not necessarily behind him" meant that he is not allowed to stand behind him (not a native speaker).
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Number of ways of arranging 7 characters in 7 spaces
Replies
8
Views
2K
Number of lines equidistant from four points on a plane
Replies
2
Views
2K
How many ways can you arrange the letters from 'GREEN' with at least one 'E'?
Replies
3
Views
2K
Finding the number of arrangements
Replies
10
Views
5K
Permutations are these not correct?
Replies
6
Views
6K
Permutation Question: Arranging 3 Cards from 6
Replies
5
Views
3K
Bishops and permutations on chessboard
Replies
23
Views
3K
What Are the Probabilities of Different Seating Arrangements at a Party?
Replies
2
Views
2K
I How Are Permutations Calculated in a Circular Seating Arrangement?
Replies
2
Views
2K
Possibly wrong permutation/combination question?
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top