anonymous
  • anonymous
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?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
It's simple, \( \large \frac{9!}{4! \times 3! \times 2!} = 1260 \)
anonymous
  • anonymous
permutation. . i'll go with FFM
anonymous
  • anonymous
Does this come from a formula? What's the rationale behind this equation?

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anonymous
  • anonymous
Also, what if order does not matter? ie, combinations?
anonymous
  • anonymous
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 \[ \large \frac{9!}{4! \times 3! \times 2!} \]
anonymous
  • anonymous
Based on your explanation, if the order does not matter, can I say it would then be this: \[\large \frac{9!}{(4! \times 3! \times 2!) \times 3!}\]
anonymous
  • anonymous
No that is incorrect interpretation for my explanation.
anonymous
  • anonymous
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.
anonymous
  • anonymous
What you mean by order does not matter, my answer assumes it means that A's and B's and C's are indistinguishable.
anonymous
  • anonymous
at xenon, i think you can only do that to combination problems, this is permutation. .
anonymous
  • anonymous
@jerwyn gayo oh yea...I think you are right. Thanks FFM for the help!! :)
anonymous
  • anonymous
Because I think if I want "order does not matter", it will just be 1.
anonymous
  • anonymous
Glad to help :)

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