Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Arbitrary Arrangement.
Let \(a_1, a_2, ....,a_{11}\) be an arbitrary arrangement (ie permutation) of the integers 1,2,....,11. Then the numbers \((a_1-1)(a_2-2)....(a_{11}-{11})\) is
Permutation
Numbers
Even and Odd
Answer: necessarily even.
B.Stat Objective Problem 119
Challenges and Thrills of Pre-College Mathematics by University Press
necessarily \(\leq\) 0 case
taking values (2-1)(3-2)(4-3).....(10-9)(1-10)(10-11)
here all the terms except last two terms are positive and there are 2 negetive terms whose product will be even
then product > 0
then not necessarily < 0 or = 0
necessarily even case
by contradiction
we assume that the product is not necessarily even
that is each of the factor have to be odd
then the arrangement look like
(even-1)(odd-2)(even-3)(odd-4)....(even-9)(odd-10)
but only one odd number left which will pair with 11 that a contradiction
\(\Rightarrow\) product is even
\(\Rightarrow\) necessarily even.
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