Probabilities?

Question

Say I have The numbers 1-2-3-4-5-6-7-8-9 How can I work out how many possisible combinations I could arrange them in?

Answer 1

362,880 possibilities By using factorials

Answer 2

Uncountably-infinite, if numbers can be repeated and the number of numbers in the combination is not limited.
1, 1-1, 1-1-1, ..., etc.

9^9 or nine to the ninth power, if numbers can be repeated but the numbers in the combination is limited to 9.


9!
That is of course if no number can be repeated, and each must be used once.
When you have a set of n numbers there will always be n! ways to arrange them.

1! = 1; 1
2! = 2; 1-2, 2-1
3! = 6; 1-2-3, 1-3-2, 2-1-3, 2-3-1, 3-1-2, 3-2-1

Another way to think about this is you have 9 total positions to fill.
x-x-x-x-x-x-x-x-x
In the first position you have all available numbers remaining to fill it. In this case {1-9}. Once the first position is filled then there are eight positions to fill and eight numbers remaining. i.e. If you use 1 in the first spot then only 2 - 9 are available to be placed in the second spot. If you continue this pattern you get 9*8*7*6*5*4*3*2*1 or 9!.

Answer 3

There are two words in probabilities that you have to be careful with. They are combinations and permutations.

Combinations means subsets of your elements where order does not count. For example (1,2,3) is the same combination
as (3,2,1) or (2,1,3) etc. The sum of all the combinations of subsets with1 digit, and 2 digits and 3 digits, ... and
all 9 digits is 2^9(the "..." ellipsis in math means you fill in the rest)
.A product determined by 9 factors of 2 which is 512.

However you used the word "arranged" which connotes
ordering to me. So the question you might be asking is
how many permutations. And since you did not specify
a subset number(how many permutations of 3 digits or 4
or 5 etc. in 9) I'll assume you mean the whole set of 9.
Now here the permutation of (1,2,3,4,5,6,7,8,9) is distinct
from (9,1,2,3,4,5,6,7,8) etc. and we calculate like this.
There are 9 different digits that could be in the 1st position.
When that position is filled there are 8 different digits that
could go in the 2nd position for a total of 9X8=72 different
sequences. Now there are 7 different digits that could go in
the 3rd position for a total 72X7=504 different sequences.
Get the idea. So the answer is
9X8X7X6X5X4X3X2X1=362,880.
We abbreviate the left side as 9! ,spoken as 9 factorial.

Answer 4

Consider there is 9 places,
First number let say "1" come in, it has 9 places to choose, so there is 9 situations for "1" choosing.
After sitting by "1", there is 8 places, second number let say "2" come in. It has remainding 8 places to choose. So there is 8 situations.
similarly, "3" has 7 places to choose, so there is 7 situations.
and so on.

So the probability is to multiply all those situations, ie
9*8*7*6*5*4*3*2*1=9!=362,880

Answer 5

9!=1 x 2 x 3 x 4 x 5 x 6x 7x 8 x9=362880

Answer 6

To arrange all of them next each other
9.8.7.6.5.4.3.2.1=9!

Answer 7

9! or Permutation(9,9)