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askme12345
 3 years ago
Best ResponseYou've already chosen the best response.0For example : 121212121, order doesnt matter how many different ways can you arrange the digets?

RagingSquirrel
 3 years ago
Best ResponseYou've already chosen the best response.0Too many to count would be my best guess.

snappyjones
 3 years ago
Best ResponseYou've already chosen the best response.19 times 9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 = 387 420 489

Hero
 3 years ago
Best ResponseYou've already chosen the best response.3Order doesn't matter and there's repetition

RagingSquirrel
 3 years ago
Best ResponseYou've already chosen the best response.0There are 1,000,000,000 different 9digit numbers, ranging from 000,000,000 to 999,999,999.

Diana
 3 years ago
Best ResponseYou've already chosen the best response.09! so it would be 9x8x7x6x5x4x3x2x1=362880

askme12345
 3 years ago
Best ResponseYou've already chosen the best response.0since there is 4 2's and 5 1's wouldnt some overlap though?

Hero
 3 years ago
Best ResponseYou've already chosen the best response.3Zarkon...any input at all??

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1are 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
 3 years ago
Best ResponseYou've already chosen the best response.3number of arrangement of 121212121. I don't remember how to do it.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0uh oh I did \[\frac{9!}{4!} + \frac{9!}{5!}\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1if you wanted to arrange 112223333 \[\frac{9!}{2!3!4!}\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1multinomial coefficients

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0what is exactly meant by order doesn't matter?

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0is 12 = 21 if order doesn't matter?

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0how would you do it for 121212121 if order matters

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1so 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
 3 years ago
Best ResponseYou've already chosen the best response.0doesn't involve nPr or nCr?

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0hmm 5 1s and 4 2s total 9 position is it \[\frac{9!}{4!} + \frac{9!}{5!}\]? (order matters)

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0I am bad at combinatorics :/
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