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Five cards are to be selected from 10 cards numbered 1 to 10 [#permalink]
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19 Nov 2009, 18:08
Question Stats:
50% (02:23) correct
50% (01:54) wrong based on 37 sessions
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Re: Really hard 800+ problem. [#permalink]
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19 Nov 2009, 21:44
Is the answer B116? I could figure out that the number of selections has to be even and less than half the total possibilities = 10c5/2 = 252/2 = 126 so B is the only possibility.
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Re: Really hard 800+ problem. [#permalink]
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19 Nov 2009, 22:18
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Bunuel wrote: Five cards are to be selected from 10 cards numbered 1 to 10. In how many ways can the cards be selected so that the average of the five numbers selected is greater than their median?
(A) 111 (B) 116 (C) 222 (D) 232 (E) 252 Number of ways to select 5 cards out of 10 cards = 10C5 = 10!/5!5! = 252 These 252 ways will have 3 possibilities: Average = Median Average > Median Average < Median These are the combinations (30 in all) where Average = Median: {1,2,3,4,5} {1,2,4,5,8} {1,2,4,6,7} {1,3,4,5,7} {2,3,4,5,6} {1,2,5,7,10} {1,2,5,8,9} {1,3,5,6,10} {1,3,5,7,9} {1,4,5,6,9} {1,4,5,7,8} {2,3,5,6,9} {2,3,5,7,8} {2,4,5,6,8} {3,4,5,6,7} {1,4,6,9,10} {1,5,6,8,10} {2,3,6,9,10} {2,4,6,8,10} {2,5,6,7,10} {2,5,6,8,9} {3,4,6,7,10} {3,4,6,8,9} {3,5,6,7,9} {4,5,6,7,8} {3,6,7,9,10} {4,5,7,9,10} {4,6,7,8,10} {5,6,7,8,9} {6,7,8,9,10}
This means there are 252  30 = 222 combinations where Average ≠ Median Now (the difficult part to explain!)  for each of the combinations where Average<Median, there will be one where Average>Median. Consider the format {1,2,3,x,x} => there are 7!/2!5! i.e. 21 ways of selecting the rest two (i.e. 21 combinations for five cards in the format 1,2,3,x,x). Out of 21 such combinations, except the one listed above where average=median, the rest 20 will have Average>Median. Now, conversely, consider the format {x,x,8,9,10} => except the one listed above where average=median, the rest 20 combinations will have Average<Median. What I am trying to say (in a longwinded way ) is that the answer should be 222/2 = 111. pls confirm the OA. Does anyone have a shorter method of solving this question? Took me a lot of time to even think of an approach on this one...



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Re: Really hard 800+ problem. [#permalink]
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19 Nov 2009, 22:40
ill agree with A 111 definitely a tricky question but if mean does not equal median, data is either skewed left or right and you can assume this is 50/50 based on randomness of selection however, determining the 30 mean=median seems like it will take a lot of time not sure if theres a shortcut for this? thanks
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Re: Really hard 800+ problem. [#permalink]
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20 Nov 2009, 07:29
This is hard!
B?
I know for sure these numbers cannot be consecutive because it they were, mean = media.
I know the number of ways to select 5 cards out of 10 cards = 10C5 = 10!/5!5! = 252
Now figure how many we can get teh average of these numbers greater than the median is the hard part. I guess B because I do some magic that give me a number close to B. So, B is my final answer.



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Re: Really hard 800+ problem. [#permalink]
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17 Feb 2010, 11:38
very interested if anyone found a way to determine the number of sets in which mean=median.
The steps to solve this problem appear to be::
1  calculate maximum # of combinations 2  subtract the number of combinations where mean=median 3  divide the remaining number by 2



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Re: Really hard 800+ problem. [#permalink]
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17 Feb 2010, 23:21
masland wrote: very interested if anyone found a way to determine the number of sets in which mean=median.
The steps to solve this problem appear to be::
1  calculate maximum # of combinations 2  subtract the number of combinations where mean=median 3  divide the remaining number by 2 mean = median in case when Difference between consecutive numbers is same i.e 1,3,5,7,9 or 1,2,3,4,5 cases. What is the OA for the question



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Re: Really hard 800+ problem. [#permalink]
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18 Feb 2010, 07:34



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Re: Really hard 800+ problem. [#permalink]
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20 Feb 2010, 11:37
Bunuel wrote: OA: A (111). Gnet's explanation is correct. +1. Hi Bunuel... do we have a shorter way for such problems?
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Re: Really hard 800+ problem. [#permalink]
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20 Feb 2010, 12:31



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Five cards are to be selected at random from 10 cards [#permalink]
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26 Sep 2012, 00:33
Five cards are to be selected at random from 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median?
a.99 b.54 c.111 d.102 e.79



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Re: Five cards are to be selected at random from 10 cards [#permalink]
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26 Sep 2012, 01:06



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Tough Combinatorics! [#permalink]
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13 Nov 2012, 21:50
Five cards are to be selected at random from 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median?
 Ans will be provided later.



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Re: Tough Combinatorics! [#permalink]
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13 Nov 2012, 22:48
Ans. 111
No. of ways of selecting 5 cards out of 10 = 10C5 = 252 ways
Now in these 252 ways there are 3 categories
1. mean = median 2. mean < median 3. mean > median
Now no. of case 2 = no . case 3
With case 1 i.e. mean = median
the sum of deviation of the nos. less than median should be equal to those greater than median
for example in case of median 3, there is 1 probability 12345 (deviations 2, 1, +1, +2) median 4, 4 probabilities 12467, 12458, 13457, 23456 (deviations [3, 2, +2, +3], [3, 2, +1, +4], [3, 1, +1, +3], [2, 1, +1, +2] similarly for median 5, there are 10 possibilities median 6, 10 ways (ssame as median 5) median 7, 4 ways (same as median 4) median 8, 1 way (same as median 3)
Hence total 30 ways
Thus the answer = (10C530)/2 = (25230)/2 = 222/2 = 111
Last edited by suryanshg on 14 Nov 2012, 01:00, edited 1 time in total.



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Re: Tough Combinatorics! [#permalink]
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13 Nov 2012, 23:35
bellcurve wrote: Five cards are to be selected at random from 10 cards numbered 1 to 10. How many ways are possible that the average of five numbers selected will be greater than the median?
 Ans will be provided later. By my logic, there is no set in which mena is lesser than the median. Either they have to be equal or the mean has to be greater. There are 10C5 ways in total. We have to subtract the number of combinations for which mean and median are equal. I'm jus stuck here though. No idea on how to get that. If answer choices were given, I'd just choose the option which is just a bit lesser than 10C5 since mode and median being equal is quite a constrained requirement and I dont think there would be that many of those. Kudos Please if my post helped.
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Re: Tough Combinatorics! [#permalink]
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14 Nov 2012, 03:04




Re: Tough Combinatorics!
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14 Nov 2012, 03:04







