30I was sure after posting that the answer was wrong. Anyway, still thinking of an approach!
There are n children on a merry-go-round. They' decide to change places so that somebody else is in front of each one. In how many ways can they achieve this?
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10 Answers
30
49I consider n=odd
first seat any one..
Every seat is identical..(consider circular arrangement)
he can be seated in 1 way.
Now one is seated in front of this man
If this person is the one who was sitting behin him..then we can seat one person behind the first person in n-2 ways.
And if the man in front was some one else..tht cud be done in n-2 ways.. and for each of these one man can be seated behind the first man in n-3 ways..
so 3 men have been seated in (n-2) + (n-2)(n-3) ways = (n-2)2 ways..
now we have a group of 3 men and we have to fill n-3 men..
now two more men will be seated in (n-3 - 2) + (n-3 -2)(n-3 - 3) = (n-5)2
Next: (n-5 -2) + (n-5 -2)(n-5 -3) = (n-7)2
next: (n-9)2
ans... (n-2)2.(n-5)2.(n-7)2......until the factor 1 or 2 is reached depending on whether the number n is even or odd..!
I am not quite sure of the answer.. but posted whatever I felt was logical.. someone please verify it! [1]
49okay..!
does in front of mean diametrically opposite??
I thought samne wala ghora! :s
1i hope it is clear enough with the phrase:
change places so that somebody else is in front of each one
isn't it?
49Someone else had posted soln wid diametrically opposite..!
And has now deleted the post..!
So I had confusion! [1]
21It confuses me..
because if u consider diametrically opposite then n must be even ......
