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How many permutations of {1, 2, 3, 4, 5, 6} exist such that: (a) No even integer is in its natural position. (b) Exactly three integers are in their natural position. (c) 1 and 6 are in their natural positions.

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  1. Yesterday, 20:22
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    (a) 6 permutations

    (b) 6 permutations

    (c) 24 permutations

    Step-by-step explanation:

    Given the set {1, 2, 3, 4, 5, 6}

    We need to check how many permutations exist such that:

    (a) No even integer is in its natural position.

    All we need to do is count the number of elements, n, between which we want to make permutations, and take factorial of n to know the number of permutations.

    The even integers are 2, 4, and 6, and we want to ensure that they are in their natural positions. So, the permutations are between 3 elements. For n = 3, factorial of 3 is 3! = 3*2*1 = 6.

    To back this up, see the permutations.

    {1, 2, 3, 4, 5, 6}, {1, 2, 5, 4, 3, 6}, {3, 2, 5, 4, 1, 6}, {3, 2, 1, 4, 5, 6}, {5, 2, 1, 4, 3, 6}, and {5, 2, 3, 4, 1, 6}

    Six permutations.

    (b) Exactly three integers are in their natural position.

    Exactly 3 integers leaves us with permutations between three numbers, so 3! = 6.

    We have six permutations.

    (c) 1 and 6 are in their natural positions

    If only 1 and 6 are in there natural position, then we have permutations between 4 elements.

    We have 4! Permutations.

    4! = 4*3*2*1 = 24
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