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Practice CAT Question

by zachlebo » Wed Jun 08, 2011 6:24 pm
Six Mobsters have arrived at the theater for the premiere of the film "Goodbuddies." one of them, 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

so N = 6!. so ince Joey cant sit in either the first spot in line or the last, you would minus out 4!? That is what I thought. i get confused on when to use permutations of combinations

b
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by zachlebo » Wed Jun 08, 2011 6:24 pm
im sorry...the correct answer is D
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by Anurag@Gurome » Wed Jun 08, 2011 6:46 pm
zachlebo wrote:Six Mobsters have arrived at the theater for the premiere of the film "Goodbuddies." one of them, 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

so N = 6!. so ince Joey cant sit in either the first spot in line or the last, you would minus out 4!? That is what I thought. i get confused on when to use permutations of combinations

b
No. of ways to arrange six mobsters = 6! = 6 * 5 * 4 * 3 * 2 = 720 ways
In 50% of 720 ways = 360 ways, Frankie will be behind Joey, and in 360 ways Joey will be behind Frankie.
So, the correct answer is D.
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by [email protected] » Wed Jun 08, 2011 7:20 pm
Can any of the experts please help in this particular question...
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by [email protected] » Wed Jun 08, 2011 7:27 pm
Anuraag could you please explain in more detail....
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by cans » Wed Jun 08, 2011 7:33 pm
6 people. F has to be behind J.
Thus F can't be on front
case 1: J at front, then rest 5 can arrange in any way -> 1*5! =120
case2: J at second: 4C1*1!*4! = 96(4C1*1! because one of the remaining person has to stand in front. After placing J, remaining 4 in any way)
case3: J at 3rd: 4C2*2!*3! = 72
case4: J at 4th: 4C3*3!*2! = 48
case5: J at 5th: 4C4*4!*1! = 24
total = 360
IMO D
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by Anurag@Gurome » Wed Jun 08, 2011 7:39 pm
[email protected] wrote:Anuraag could you please explain in more detail....
When we're arranging n distinct objects, there are n! possible arrangements.

So, total # of arrangements = 6! = 6 * 5 * 4 * 3 * 2 = 720.

Frankie, insists upon standing behind Joey in line at the concession stand.

There are 2 possibilities, either Frankie will be behind Joey or Joey will be behind Frankie, and either of this scenario is equally likely. Frankie will be behind Joey 50% 0f the times. And we have already found that total # of arrangements = 720. So, Frankie will be behind Joey 50% of the times implies 50% of 720 = 360.

Hope that helps.
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by peelamedu » Wed Jun 08, 2011 7:59 pm
Awesome way to solve Anurag. Thanks.
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by [email protected] » Wed Jun 08, 2011 10:09 pm
Great explanation by both ... Thank You!!!
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by zachlebo » Thu Jun 09, 2011 7:27 am
hypothetically, if the question said that they had to be DIRECTLY next to each other, would the answer change?

ps. thanks for the responses! i appreciate it!!
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