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Thread: Help with this permutation problem?

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    Help with this permutation problem?




    Hi! So I'm struggling with a lot of things in probability, and one being permutations and finding them.

    There's a problem, and if I know how to do it I'll be good for the test on Monday.

    The problem is this:

    - The 11 digits 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4 are permuted in all distinguishable ways. How many permutations (a) begin with 22? (b) begin with 343?

    So I'm kinda lost here, any help would be appreciated. I tried this:

    10[(9!)/((1!)(1!)(4!)(3!)] and then divided [(11!)/((4!)(3!)(3!)(1!))] from it and got .545454545, I do not think that's right lol. So I guess I'm fuzzy on how this all works.

    My professor gave us a similar yet simpler example. If all 11 letters in the word "Mississippi" were scrambled, how many permutations can result?

    Well he showed that the answer is 11 choose 4, 4, 2 and 1. But how is this calculated? Is it (11!)/((4!)(4!)(2!)(1!))? Or something else?

    Thanks!!



    Problem from:

    Goldberg, Samuel (2013-04-22). Probability: An Introduction (Dover Books on Mathematics) (Kindle Locations 3796-3797). Dover Publications. Kindle Edition.

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    Re: Help with this permutation problem?

    hi,
    if you have n elements, out of which k1, k2..etc are identical then the number of permutations that are distinguishable will be n!/(k1!k2!..). So, if you fix 22 as the starting sequence you just need to count how many arrangements you have of the remaining elements. Same for the second problem imo.

    i hope this helps, it has been a while since I solved combinatorial problems, so check if this makes sense for you.

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    Re: Help with this permutation problem?

    That does help a bit, yeah. So the answer for a would be: (9!)/(1!1!3!4!) ? And B would be: (8!)/(1!3!2!3!)?

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    Re: Help with this permutation problem?


    I think, yes, except by B you only have two 4s left

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