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Question
Let the eleven letters, A,B,....K denote an artbitrary permutation of the integers (1,2,....11), then (A−1)(B−2)(C−3)...(K−11) is
Let the eleven letters, A,B,....K denote an artbitrary permutation of the integers (1,2,....11), then (A−1)(B−2)(C−3)...(K−11) is
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Solution
The correct option is C Always evem
Given set of numbers is {1,2,3.....11} in which 5 are even and six are odd, which demands that in the given product it is not possible to arrange to subtract only even number from odd numbers. There must be at least one factor involving subtraction of an odd number from another odd number. So at least one of the factors is even. Hence product is always even.
Given set of numbers is {1,2,3.....11} in which 5 are even and six are odd, which demands that in the given product it is not possible to arrange to subtract only even number from odd numbers. There must be at least one factor involving subtraction of an odd number from another odd number. So at least one of the factors is even. Hence product is always even.
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