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Counting - Six mobsters

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by Taran » Mon Dec 13, 2010 10:26 pm
@sanjana - You did it the right way. You however made an error while adding up the different combinations. :)
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by gmat3last » Tue Mar 22, 2011 5:59 am
HELP...I know this is a little off-topic, but I could not find think of a better place to discuss a math problem (since am a newbie)

Question: How many different positive divisors are there to 2600?
Can anyone show me the quickest way to solve this?
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by Aditya57 » Thu May 26, 2011 10:28 am
Sanjana ,your reasoning is right and sum it up in the manner, 120+96+72+48+24 equals 360 only ,not 350 B-) and my reasoning goes as
5! for Frankie standing last in line
4*4! for Frankie being 5th in line
3*4! for Frankie being 4th in line
2*4! for being 3rd in line
1*4! for being 2nd in line
-----
4! for the four positions other than Frankie and Joey in line ,remaning numbers for the number of positions Joey can stand in the line ,
so (4+3+2+1)*4! + 5!= 360
and simpler one is 6!/2 ,a 50-50 chance ! Thanks
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by Aditya57 » Thu May 26, 2011 10:36 am
gmat3last wrote:HELP...I know this is a little off-topic, but I could not find think of a better place to discuss a math problem (since am a newbie)

Question: How many different positive divisors are there to 2600?
Can anyone show me the quickest way to solve this?
---------------------------
Hey it's a simple one take the number given,
1. express the number as product prime factors.
2. exponent the common factor and writing the prime factors as powers.
3. add 1 to number of times each prime factor is present.
4. multiply them and that is your answer.
ex: 2600
can be expressed as 2*2*2*5*5*13
= (2^3)*(5^2)*(13^1)
=(3+1)*(2+1)*(1+1)
=4*3*2
= 24 divisors
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by saketk » Thu Aug 25, 2011 8:05 pm
Starting with the first position-- If frankie is in 1st pos.. rest 5 can arrange themselves in 5! ways...
likewise there will be 4 more case-- and the resultant of these cases will be

4! (1+2+3+4) = 240

Total = 240+5! = 240+120 = 360 :)
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by quantskillsgmat » Thu Aug 25, 2011 9:49 pm
i think problem is same as
in how many ways abcdef can be arranged so that a is always ahead of b.(there may be zero or more persons between a and b.)
answer is 360
becz when n different thing arranged in a line theny can be arranged in n factorial ways.and every element has half of the cases before it and half behind it.
so desired answer is 6 factorial/2=360
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by quantskillsgmat » Thu Aug 25, 2011 10:42 pm
this was easy questionwhat else. if conditions are something different
eg
abcdef is to be arranged in a line in how any ways it can be done if a is ahead of b and b is ahead of c.
i think answer shd be 6!/2!x2!
am i ryt bent
thanx
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by olegpoi » Fri Aug 26, 2011 1:51 pm
360

There are 5 positions in the stand for F to keep his eye on J:
6th 5*4*3*2*1x1=120
5th 4x4*3*2x1x1=96
4th 3x4*3x1x2*1=72
3rd 2x4x1x3*2*1=48
2nd 1x1x4*3*2*1=24
Total of scenarios =360
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by Deependra1 » Sun Sep 04, 2011 3:51 am
ANSWER: D
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by knight247 » Sun Sep 04, 2011 1:14 pm
I kinda tried solving this using only elimination.

If no restrictions then they can be arranged in 6! ways=720 so that eliminates E

If Frank is always behind Joe then we consider them as one. I'm writing them as JF and the other mobsters as M1 M2 etc

JF M1 M2 M3 M4
These can be arranged in 5! ways =120. However there are other permutations to be considered as well. So the actual figure is much greater than 120. So that eliminates A, B and C. Hence D
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by prashant misra » Thu Sep 08, 2011 9:01 am
my oficial answer is also 360.thank god i did the question through my logic
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by jsnipes » Wed Sep 21, 2011 6:32 pm
i got the right answer but i think i may have done it wrong. so total number of ways that the 6 people can be arranged (since order matters?) is 6! or 720. i then took 720 and divided it by the number of people who have to stand together as 2! not two. so i got 6!/2! which for 2 is obviously the same thing and results in the correct answer of 360. if there were three people would I divide 720 by 3 or 3!

thanks for help, i know this is two day old question i just got around to answering it though.
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by parul9 » Thu Sep 29, 2011 9:46 am
I got he answer right - 360.
but I solved it the long way...
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by him1985 » Sun Jan 29, 2012 3:36 am
Good question indeed!!
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
Himanshu Chauhan
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by ronnie1985 » Sun Jan 29, 2012 10:44 am
The shortcut is very difficult to strike at exam time. the long method although long takes short time for calculation. I would have gone for the longer method rather than short one during exam as it yields accuracy and confidence.
Follow your passion, Success as perceived by others shall follow you
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