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Manhattan GMAT challenge question

by gabriel » Wed Sep 26, 2007 7:43 am
Source : - Manhattan GMAT (www.manhattangmat.com)

Manhattan GMAT’s football team has 99 players. Each player has a uniform number from 1 to 99 and no two players share the same number. When football practice ends, all the players run off the field one-by-one in a completely random manner. What is the probability that the first four players off the field will leave in order of increasing uniform numbers (e.g., #2, then #6, then #67, then #72, etc)?

(A) 1/64
(B) 1/48
(C) 1/36
(D) 1/24
(E) 1/16

Try it guys ..
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by Nisha1218 » Wed Sep 26, 2007 8:32 am
Is the answer D?
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by gabriel » Wed Sep 26, 2007 8:46 am
Nisha1218 wrote:Is the answer D?
I have no clue what the answer is ... will post my answer after a few more replies ..

Btw care to post ur method too
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by Nisha1218 » Wed Sep 26, 2007 9:27 am
the method i used to solve it was to add the following

1/99 (#2)

1/98 (#6, also decreasing the pool to 98 since 2 is no longer there)

1/97 (#67, also decreasing the pool to 98 since 2 & 6 are no longer there)

1/96 (#72, also decreasing the pool to 98 since 2,6, & 67 are no longer there)

= 1/24
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by raulverde » Wed Sep 26, 2007 9:52 am
Nisha1218,

That was just an example : it could 1,3,6,7 as well . or any other increasing order
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by Nisha1218 » Wed Sep 26, 2007 10:05 am
True, but then could the probabilities change?
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by raulverde » Wed Sep 26, 2007 10:21 am
Yes they would....for example you pick 2 as the first number then for the next number you can pick any number from 3 to 97. And so on...
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by samirpandeyit62 » Wed Sep 26, 2007 11:06 am
IMO, the probabilties of every player will depend upon its previous players number, so there can multiple probabilities for 4 players coming out in increasing order, depending upon the shirt number players who come out, so I think we need to evaluate the worst scenario here which will be if player 96 comes out first

i.e probalilty of 96 =1/99

then 97 should come out i.e P =1/98

then 98 should come out wih P =1/97

and lastly 99 with P =1/96

if we multilply these we should get the reqd probabilty, however it would not be one amongst the answer chioces.

Gabriel what are u thoughts
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by gabriel » Wed Sep 26, 2007 12:12 pm
My answer is 1/24 .. here's how i got it ..

definition of probability = favorable conditions/ all possible conditions ..

now, all possible conditions = 99p4 .. it will be permutations over here because sequence matters ..

Now, comes the interesting part, favorable conditions .. Consider any 4 numbers eg, 1, 10, 20, 25 ... these numbers can be arranged among themselves in 4! ways .. but out of these arrangements only one will have the numbers in ascending order .. this can be proved by choosing any 4 numbers .. So, for every group of 4 there is only one way our condition can be met .. so our question basically boils down to, how many groups of 4 can we get by randomly selecting 4 out of 99, and the answer is 99c4 ..

So the probability is 99c4/99p4 = 1/24 ..


PS: - I am not sure about the answer, the answer is only known to Manhattan GMAT students, if we have one of them among us, then please do post the OA
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by camitava » Wed Sep 26, 2007 9:04 pm
Gabriel dnt mind! But I have not understood one thing. The question says about a football team and so there should 11 players. But why you people (I mean Samir, Nisha and you) refer to 4 at all i.e 6C4 or 9P4? As per my understanding -
11 Players are having 99 T-shirts with them
They will come to ground wearing the T-shirts in a fashion that difference between the number of two players are in increasing order. [E.g. - 1, 2, 4, 7, 11 and so on
Or
2, 3, 5, 8, 12, 17 and so on
Or ...]

If so,
U can arrange the possibilities as 99P11. Right?
And for the desired outcome the possibilities are -
[Initial difference between two players can not go beyond 5. And if the difference is 5, then starting number should be 1. So with the starting number as 1, we can have max 5 different combinations. For two, it will be something like 5 again and so on until the starting number is 5. So from 1 to 5 as starting number, we are having 5 different possibilities for for each.
So the possibilities for desired outcome is - 25. So the probability is 25/99P4. Yop, this answer is not provided in any of the options but still ... ]
Looking for ur responses !!! :wink: :)
Correct me If I am wrong


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Amitava
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by samirpandeyit62 » Wed Sep 26, 2007 10:15 pm
Hey Amitava,
Man u have used some critical reasoning here :D ,I think u will crack CR's :D , however I dont think we can assume any thing other than what is given in the question, I agree with gabriel the answer should be 1/24

99P4 = all possibilities(orders) of 4 players coming out
99C4 = only one order of any 4 players coming out, which corresponds to the 1 order where the players come out in ascending order.

by the way gabriel this was a very good question man, post more like these one's, they would really help in enhancing the thinking spectrum of everyone.

also gabriel sometimes after seeing such problems I think to study P&C's of the CAT level. What do say abt this. :?:

The soln part here will take only 10-20 secs, but the thinking part may take much more than 2 mins with total unfamiliarity.
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by camitava » Wed Sep 26, 2007 10:23 pm
Samir, I am not getting why u people are taking 4 people at a time-that's why u r using 99C4 or 99P4. Where from u r gettin the number - 4? A football team (mentioned in the qs itself), generally consists of 11 players. Can u pls explain me why u r taking 4 as a number?
Correct me If I am wrong


Regards,

Amitava
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by samirpandeyit62 » Wed Sep 26, 2007 11:08 pm
If u refer the question stem, it says

"What is the probability that the first four players off the field will leave in order of increasing uniform numbers (e.g., #2, then #6, then #67, then #72, etc)? "

So we need to find the probability of first 4 players in ascending order, for which we need all the permutations of 4 players selected out of 99, this would be total nos of possibilities, & for each group of 4 nos there would be only one order in which the are in ascending order, which corresponds to the Combination of the 4.

Now we cannot assume 11 players in a team coz this is mentioned as a pratice session where all the 99 players might be practicing together or in individual teams of 11.
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by camitava » Thu Sep 27, 2007 12:25 am
Sorry Samir! My mistake. I admit that I overlooked that. Thanks for pointing my fault. Thanks once again!!!
Correct me If I am wrong


Regards,

Amitava
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by samirpandeyit62 » Thu Sep 27, 2007 12:34 am
Hey Amitava,
No probs man, but I must say that this problem was a gold pack of P&C & probabilty, very simple if one's concepts are 100% clear, What do u say,and really good for benchmarking ur skills,

Thanks gabriel for posting this one.
Regards
Samir
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