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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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

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??

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

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 :-/