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askme12345Best ResponseYou've already chosen the best response.0
For example : 121212121, order doesnt matter how many different ways can you arrange the digets?
 2 years ago

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

snappyjonesBest ResponseYou've already chosen the best response.1
9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 = 387 420 489
 2 years ago

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

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

DianaBest ResponseYou've already chosen the best response.0
9! so it would be 9x8x7x6x5x4x3x2x1=362880
 2 years ago

askme12345Best ResponseYou've already chosen the best response.0
since there is 4 2's and 5 1's wouldnt some overlap though?
 2 years ago

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

ZarkonBest ResponseYou've already chosen the best response.1
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
 2 years ago

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

Ishaan94Best ResponseYou've already chosen the best response.0
uh oh I did \[\frac{9!}{4!} + \frac{9!}{5!}\]
 2 years ago

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

ZarkonBest ResponseYou've already chosen the best response.1
multinomial coefficients
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
what is exactly meant by order doesn't matter?
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
is 12 = 21 if order doesn't matter?
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
how would you do it for 121212121 if order matters
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.1
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.
 2 years ago

pokemon23Best ResponseYou've already chosen the best response.0
doesn't involve nPr or nCr?
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
hmm 5 1s and 4 2s total 9 position is it \[\frac{9!}{4!} + \frac{9!}{5!}\]? (order matters)
 2 years ago

Ishaan94Best ResponseYou've already chosen the best response.0
I am bad at combinatorics :/
 2 years ago
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