71Since , no person has the same height, so If we divide them into two groups, they can be arranged in only one way (Presumably).
So 2nCn ?
I n how many ways is it possible to arrange 2n persons of different height in two ranks of n persons each so that in each rank they stand according to height and, besides, so that each man in the first rank is taller than the man behind him in the second rank?
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9 Answers
1isn't this a classic?but sadly i got only the question not the solution
71
1for any selection of n out of the 2n people, there exists just one way of such an arrangement
hence 2nCn
30This question went no where. Reviving the post, any idea what the answer could be?
341I checked with a Romanian teacher who I know from AOPS and i found that Nishant sir is bang on target.
From the input I got from him, this is known as the standard Young's 2Xn tableau where every row and every column is increasing
I am yet to read up more on this, but the guy also mentioned that this is a special case of the hook-length formula.
after this google is your friend
62This question reminded me of an old question where there was a queue where there were 2n people...
n of them had 10 rupee notes and n ahd 5 rupeee..
The cost of the ticket was 5.
The ticket counter person did not have any money at all...
Find the number of ways the people could stand in a queue so that the counter person could return them the money
@Prophet sir, this application is direct from the grid question also that we have discussed here... Where we have to go from one corner of a square grid to another without crossing the diagonal....
A little thought will prove that these are the same cases [1]
To be strictly speaking.. Those appearing for IIT JEE need to worry.