This is a little bit easier than the previous Induction qn:
The problem may be familiar to some - it is from an old Hungarian Olympiad
(a1,a2,a3,...,an) is a permutation of (1,2,3,...,n). n is an odd integer
Prove that (a1-1) (a2-2)...(an-n) is even
The well known solution is to see that (a1-1)+(a2-2)+...+(an-n) = 0 which implies that at least one of the terms has to be even.
But the proof using induction is also interesting. (and I havent come across it)
9P(n+2) = P(n) . (an+1 - n+1)(an+2 - n+2)(n+1 -r1)(n+2 - r2) /(an+1-r2)(an+2- r1)
when an+1 ≠n+1 and so on
using above statement and some logic i can prove by induction
but is this the type of proof u r lookin sir ?
or is there a more elegant MI involved
341It seems to be the lines on which I thought. but I have a request:
The point of my posting the question is that someone can pick up a line of thinking from the solution. So please post a full solution.
9ok sir
let P(n) be some permuation for odd n
------------------------------------------------------------------------------------------------------------------
P(n+2) =P(n) (a n+1 - n+1)(a n+2 - n+2)(n+1 -r 1)(n+2 - r 2) /(a n+1-r 2)(a n+2- r 1)
with a n+1 = r2 +2λ+1 , an+2= r1 +2β+1 ..........[ 1 ]-----------------------------------------------------------------------------------------------------------------
its obvious that if a n+1 = r2 +2λ or an+2= r1 +2λ
P(n+2) will be even [ P(n+2) = ....... (a n+1 - n+1)(n+2 - r 2) if a n+1 = r2 + 2λ then the expression in bold will be even ,similar argument for the other case]
now
for the other case
we see that in [ 1 ] the denominators are both odd
so whenever P(n) is even P(n+2) is also
as P(1) is true
hence P(n) is true for all odd n
341nice. thanks a lot for your patience in typing out the whole solution.
maybe the post should be shaded a darker pink :D
9yes i m doing this in my holidays only
so it wont be a problem for me as long as it helps others :)