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askme12345

  • 3 years ago

How many different ways can a nine digit number be written?

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  1. askme12345
    • 3 years ago
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    For example : 121212121, order doesnt matter how many different ways can you arrange the digets?

  2. RagingSquirrel
    • 3 years ago
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    Too many to count would be my best guess.

  3. snappyjones
    • 3 years ago
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    9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 = 387 420 489

  4. Hero
    • 3 years ago
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    Order doesn't matter and there's repetition

  5. RagingSquirrel
    • 3 years ago
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    There are 1,000,000,000 different 9-digit numbers, ranging from 000,000,000 to 999,999,999.

  6. pokemon23
    • 3 years ago
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    grrr

  7. Hero
    • 3 years ago
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    That doesn't really help

  8. Diana
    • 3 years ago
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    9! so it would be 9x8x7x6x5x4x3x2x1=362880

  9. askme12345
    • 3 years ago
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    since there is 4 2's and 5 1's wouldnt some overlap though?

  10. Diana
    • 3 years ago
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    no it shouldnt

  11. Hero
    • 3 years ago
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    Zarkon...any input at all??

  12. Zarkon
    • 3 years ago
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    are you looking to find the number of arrangement of 121212121 or any 9 digit number. The first is trivial...the second is a much longer computation

  13. Hero
    • 3 years ago
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    number of arrangement of 121212121. I don't remember how to do it.

  14. Zarkon
    • 3 years ago
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    \[\frac{9!}{5!4!}\]

  15. Ishaan94
    • 3 years ago
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    uh oh I did \[\frac{9!}{4!} + \frac{9!}{5!}\]

  16. Zarkon
    • 3 years ago
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    if you wanted to arrange 112223333 \[\frac{9!}{2!3!4!}\]

  17. Zarkon
    • 3 years ago
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    multinomial coefficients

  18. Ishaan94
    • 3 years ago
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    what is exactly meant by order doesn't matter?

  19. Ishaan94
    • 3 years ago
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    is 12 = 21 if order doesn't matter?

  20. Zarkon
    • 3 years ago
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    yes

  21. Ishaan94
    • 3 years ago
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    how would you do it for 121212121 if order matters

  22. Zarkon
    • 3 years ago
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    so if you didn't care about the order of 12121212 you would treat all the \(\frac{9!}{5!4!}\) ways to arrange it as being the same.

  23. pokemon23
    • 3 years ago
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    doesn't involve nPr or nCr?

  24. Ishaan94
    • 3 years ago
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    hmm 5 1s and 4 2s total 9 position is it \[\frac{9!}{4!} + \frac{9!}{5!}\]? (order matters)

  25. Ishaan94
    • 3 years ago
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    I am bad at combinatorics :-/

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