askme12345
How many different ways can a nine digit number be written?



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askme12345
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For example : 121212121, order doesnt matter how many different ways can you arrange the digets?

RagingSquirrel
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Too many to count would be my best guess.

snappyjones
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9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 =
387 420 489

Hero
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Order doesn't matter and there's repetition

RagingSquirrel
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There are 1,000,000,000 different 9digit numbers, ranging from 000,000,000 to 999,999,999.

pokemon23
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grrr

Hero
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That doesn't really help

Diana
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9! so it would be 9x8x7x6x5x4x3x2x1=362880

askme12345
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since there is 4 2's and 5 1's wouldnt some overlap though?

Diana
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no it shouldnt

Hero
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Zarkon...any input at all??

Zarkon
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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

Hero
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number of arrangement of 121212121. I don't remember how to do it.

Zarkon
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\[\frac{9!}{5!4!}\]

Ishaan94
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uh oh I did \[\frac{9!}{4!} + \frac{9!}{5!}\]

Zarkon
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if you wanted to arrange 112223333
\[\frac{9!}{2!3!4!}\]

Zarkon
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multinomial coefficients

Ishaan94
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what is exactly meant by order doesn't matter?

Ishaan94
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is 12 = 21 if order doesn't matter?

Zarkon
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yes

Ishaan94
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how would you do it for 121212121 if order matters

Zarkon
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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.

pokemon23
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doesn't involve nPr or nCr?

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

Ishaan94
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I am bad at combinatorics :/