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Math Help - n > 1, prove the number of even perm. of [n] equals the number of odd perm. of [n]

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    n > 1, prove the number of even perm. of [n] equals the number of odd perm. of [n]

    For n > 1, prove that the number of even permutations of [n] equals the number of odd permutations of [n].
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    Quote Originally Posted by qtpipi View Post
    For n > 1, prove that the number of even permutations of [n] equals the number of odd permutations of [n].
    Let \pi be a permutation of [n]. Define another permutation \sigma by switching \pi(1) and \pi(2); i.e., define

    \sigma(1) = \pi(2)
    \sigma(2) = \pi(1)
    \sigma(j) = \pi(j) for j > 2.

    Since \pi and \sigma differ only by a transposition, they have opposite parity, i.e. one is odd and one is even. So we have established a bijection between the even and odd permutations, and there are equal numbers of each.
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