Problem Solving

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?
1) 6
2) 24
3) 120
4) 360
5) 720

I do not understand the solution provided by MGMAT where they mention that "each of those 720 arrangements, Frankie must be either ahead of or behind Joey". Now they have this constraint that Frankie will always be behind Joey so how come he must be either ahead or behind Joey.

Can someone please explain this in some other way.

Thanks.

MGMAT explanation

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.

abhi75,
At first, I tried to solve this problem by taking diff situations in to account and tried to take every possible outcome. But this approach seemed to me little lenghty and tiresome. After that, I went to go thru Manhattan's approach. And they have solved the problem in a very easy approach. Now what I have understood from their approach is -
We can arrange 6 people without any constraint in 6! ways = 720 ways. Now in theses cases, there will be situations where -
1. J is standing before F. -> Desirable according to the Qs.
2. J is standing after F. -> Not desirable.
Only two cases can be there. So our desired cases can be in 720/2 = 260 ways. Got my response, abhi75? Pls let me know ...

thanks camitava. Now I understand they took 2 possibilities 1) in which joey will be in front and one in which he wont be.

-a

thanks.

I see why solving it the above way is best.

But I solved it this way:

If frankie is the front, there are 5 spots behind him for Joey
If frankie is 2nd from front, there are 4 spots behind him for joey
Total, there are 5+4+3+2+1 = 15 different combinations of Frankie / Joey locations

Now, there are 4 spots left in each scenario, with 4 people to fill them. 4C4 = 24.

15 x 24 = 360

Something does not appear to be correct to me. 4C4 is not 24, 4C4 = 1. Seems to me we need to use permutations here. The remaining 4 mobsters can be arranged in 4 different ways over those remaining 4 open positions in 4! = 24 ways. Rest seems ok...

Something does not appear to be correct to me. 4C4 is not 24, 4C4 = 1. Seems to me we need to use permutations here. The remaining 4 mobsters can be arranged in 4 different ways over those remaining 4 open positions in 4! = 24 ways. Rest seems ok...

Yes we need to use permutation since A behind B and B behind A are different arrangements

Let calculate the number of arrangemtnts without any constraints =

6 5 4 3 2 1 = 720 (1) (6 people to choose for first spot,5 in the next spot etc...)

Lets calculate the arrangements where Frankie is in front of Joey =

(FRANKIE IN 1ST SPOT)

F 5 4 3 2 1 = 120

(F IN 2NSD SPOT)

4 F 4 3 2 1 = 96

(F IN 3RD SPOT)

4 3 F 3 2 1 = 72

(F IN 4TH SPOT)

4 3 2 F 2 1 = 48

(F IN 5TH SPOT)

4 3 2 1 F J = 24

Add all u will get 360 (2)

No of of arrangments where Frankie is behind Joey = (1) - (2)

= 720-360
= 360

Ohwell, hope this helps let me know if u still hv questions Bud!

I did not have any question, I was just checking the answer of Cornell2005 and I noticed the minor error that I tried to correct. Thanks cramya for looking out for me though

Anytime my friend; we are all here to help each other out in the process!
Good luck!