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Math Help - Permutations.

  1. #1
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    Permutations.

    9 cards are selected with replacement from a standard pack of 52 cards with 12 picture, 20 odd and 20 even cards.

    a)How many different sequences of 9 cards are possible?
    b)How many of the sequences in part (a) will contain 4 picture cards,3 odd and 2 even cards?

    For part a) it will be 9 spaces all of which have 52 possibilities, so 52^9.
    For part b) it will be 12^4 x 20^3 x 20^2. But the picture cards can be arranged so that the picture isnt just in the first 3 positions and the odd and even to be in other positions also.
    So, the number of sequences will be 12^4 x 20^3 x 20^2 x 9!/(4!3!2!).

    If there is a mistake or you have something to add to the logic I used, please help.
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  2. #2
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    Permutations

    Hello woollybull
    Quote Originally Posted by woollybull View Post
    9 cards are selected with replacement from a standard pack of 52 cards with 12 picture, 20 odd and 20 even cards.

    a)How many different sequences of 9 cards are possible?
    b)How many of the sequences in part (a) will contain 4 picture cards,3 odd and 2 even cards?

    For part a) it will be 9 spaces all of which have 52 possibilities, so 52^9.
    For part b) it will be 12^4 x 20^3 x 20^2. But the picture cards can be arranged so that the picture isnt just in the first 3 positions and the odd and even to be in other positions also.
    So, the number of sequences will be 12^4 x 20^3 x 20^2 x 9!/(4!3!2!).

    If there is a mistake or you have something to add to the logic I used, please help.
    Your answers look good to me.

    For part (b) an alternative way of arriving at the answer would be:

    Choose the picture cards in 12^4 ways and the positions they occupy in ^9C_4 ways. Then choose the odd cards in 20^3 ways, and their positions in ^5C_3 ways. Finally choose the even cards in 20^2 ways for the last two positions.

    Total number of ways: 12^4 \times ^9C_4 \times 20^3 \times ^5C_3 \times 20^2

    = 12^4 \times \frac{9!}{5!4!} \times 20^3 \times \frac{5!}{3!2!} \times 20^2

    = \frac{12^4 \times 20^3 \times 20^2 \times 9!}{4!3!2!}

    Grandad
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