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xEnOnn Group Title

Suppose I have the following sequence in an event: BABCCABAA There are 9 elements in the sequence and I want to find the number of arrangements I get can out of this 9 elements. The order does matter but because there are repeated elements such as 4A's, 3B's, etc, it becomes not as easy as just 9 factorial. For example, the following 2 are the considered only one arrangement: \[BA_3BCCA_4BA_1A_2\] and \[BA_1BCCA_4BA_2A_3\] So how can I find the number of arrangements when there are such repeated elements in it?

  • 2 years ago
  • 2 years ago

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  1. FoolForMath Group Title
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    It's simple, \( \large \frac{9!}{4! \times 3! \times 2!} = 1260 \)

    • 2 years ago
  2. jerwyn_gayo Group Title
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    permutation. . i'll go with FFM

    • 2 years ago
  3. xEnOnn Group Title
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    Does this come from a formula? What's the rationale behind this equation?

    • 2 years ago
  4. xEnOnn Group Title
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    Also, what if order does not matter? ie, combinations?

    • 2 years ago
  5. FoolForMath Group Title
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    Put separate stickers on each of the four 'A', three 'B', and one of the two C's, to distinguish them. Now there are 9 distinct letters can be arranged in 9! ways. Remove the stickers, of all A's, then all B's and then all C's. Each time we remove the stickers, 4! arrangements collapse into 1 for A's, 3! arrangement collapses to 1 for B's and 2! to 1 for C. So the number of arrangements the given 9 letters is \[ \large \frac{9!}{4! \times 3! \times 2!} \]

    • 2 years ago
  6. xEnOnn Group Title
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    Based on your explanation, if the order does not matter, can I say it would then be this: \[\large \frac{9!}{(4! \times 3! \times 2!) \times 3!}\]

    • 2 years ago
  7. FoolForMath Group Title
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    No that is incorrect interpretation for my explanation.

    • 2 years ago
  8. xEnOnn Group Title
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    But if the order does not matter, then I need to divide away the number of stickers I remove too, right? Although I remove 9 stickers in total, I can't divide by 9! again. Otherwise the equation will become a fraction.

    • 2 years ago
  9. FoolForMath Group Title
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    What you mean by order does not matter, my answer assumes it means that A's and B's and C's are indistinguishable.

    • 2 years ago
  10. jerwyn_gayo Group Title
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    at xenon, i think you can only do that to combination problems, this is permutation. .

    • 2 years ago
  11. xEnOnn Group Title
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    @jerwyn gayo oh yea...I think you are right. Thanks FFM for the help!! :)

    • 2 years ago
  12. xEnOnn Group Title
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    Because I think if I want "order does not matter", it will just be 1.

    • 2 years ago
  13. FoolForMath Group Title
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    Glad to help :)

    • 2 years ago
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