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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

- necessarily \(\leq\) 0
- necessarily even
- necessarily 0
- none of these

Permutation

Numbers

Even and Odd

But try the problem first...

Answer: necessarily even.

Source

Suggested Reading

B.Stat Objective Problem 119

Challenges and Thrills of Pre-College Mathematics by University Press

First hint

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

Second Hint

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

Final Step

\(\Rightarrow\) product is even

\(\Rightarrow\) necessarily even.

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