Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

For example : 121212121, order doesnt matter how many different ways can you arrange the digets?
Too many to count would be my best guess.
9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 times 9 = 387 420 489

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question