askme12345
  • askme12345
How many different ways can a nine digit number be written?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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askme12345
  • askme12345
For example : 121212121, order doesnt matter how many different ways can you arrange the digets?
RagingSquirrel
  • RagingSquirrel
Too many to count would be my best guess.
anonymous
  • anonymous
9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 = 387 420 489

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More answers

Hero
  • Hero
Order doesn't matter and there's repetition
RagingSquirrel
  • RagingSquirrel
There are 1,000,000,000 different 9-digit numbers, ranging from 000,000,000 to 999,999,999.
pokemon23
  • pokemon23
grrr
Hero
  • Hero
That doesn't really help
Diana
  • Diana
9! so it would be 9x8x7x6x5x4x3x2x1=362880
askme12345
  • askme12345
since there is 4 2's and 5 1's wouldnt some overlap though?
Diana
  • Diana
no it shouldnt
Hero
  • Hero
Zarkon...any input at all??
Zarkon
  • Zarkon
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
Hero
  • Hero
number of arrangement of 121212121. I don't remember how to do it.
Zarkon
  • Zarkon
\[\frac{9!}{5!4!}\]
anonymous
  • anonymous
uh oh I did \[\frac{9!}{4!} + \frac{9!}{5!}\]
Zarkon
  • Zarkon
if you wanted to arrange 112223333 \[\frac{9!}{2!3!4!}\]
Zarkon
  • Zarkon
multinomial coefficients
anonymous
  • anonymous
what is exactly meant by order doesn't matter?
anonymous
  • anonymous
is 12 = 21 if order doesn't matter?
Zarkon
  • Zarkon
yes
anonymous
  • anonymous
how would you do it for 121212121 if order matters
Zarkon
  • Zarkon
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.
pokemon23
  • pokemon23
doesn't involve nPr or nCr?
anonymous
  • anonymous
hmm 5 1s and 4 2s total 9 position is it \[\frac{9!}{4!} + \frac{9!}{5!}\]? (order matters)
anonymous
  • anonymous
I am bad at combinatorics :-/

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