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?
And how do I do the Equation?
2006-09-30 07:04:40 · 11 answers · asked by ProfessorCarrot 1 in Science & Mathematics ➔ Mathematics
Just use the factorial 9! That means 9*8*7*6*5*4*3*2*1 which is 362,880
2006-09-30 07:08:10 · answer #1 · answered by MateoFalcone 4 · 0⤊ 0⤋
This is part of iteration you basically raise the base to the power
You have 1 to 9
This is a base 9 number system.
If you want to know how many three digit numbers you can make from them then you raise 9 [the base] to the power of 3 [the number of places.
9 ^ 3 = 729 combos
This is because we hav 9 numbers for place 1, 9 for place 2 and 9 for place 3 this is equal to 9 * 9 * 9
If you want to know how many orders you can place 1 to 9 in a nine digit grouping using each number only once then you simply do:
9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362,880
This is because if we only use each number once then we have 9 numbers for place 1, 8 for place 2, 7 for place 3 and so on
I apologise if this is confusing but it is complicated so i would advise reading it a few times and even trying it [with less numbers in your base]
2006-09-30 07:24:55 · answer #2 · answered by Anonymous · 0⤊ 0⤋
That's 9 factorial = 9*8*7^6*5*4*3*2*1
That means you have 9 numbers to choose from initially. After using a number, you have 8 numbers to choose from... then you have 7, then 6..... until you you are left with only one number.
2006-09-30 07:14:58 · answer #3 · answered by Anonymous · 0⤊ 0⤋
You may want to add more details to your question so that we'll know what you want to know.
- If you mean that all possible combinations of the numbers you have given without any restrictions - numbers can be repeated, any number of digit (1-9), any order then it would be
C = 9C1 + 9C2 + 9C3 + 9C4 + 9C5 + 9C6 + 9C7 + 9C8 + 9C9
- If you mean that all combinations without repetition and only 9 digits, then it would be:
C = 9*8*7*6*5*4*3*2*1 = 9!
* There's a lot more combinations so you may want to explain your question a bit further
2006-09-30 10:16:48 · answer #4 · answered by augel 2 · 0⤊ 0⤋
9! (i,e, 9 factorial = 9*8*7*6*5*4*3*2*1)
2006-10-03 12:06:50 · answer #5 · answered by vish 2 · 0⤊ 0⤋
2^9
2006-09-30 07:07:07 · answer #6 · answered by iyiogrenci 6 · 0⤊ 2⤋
it depends on how you want to arrange? Does order matter? or do you just need the all possible combination just use the 9!
For other answers use: n!/r! or n!/r!(n-r)! formulas
2006-09-30 07:21:33 · answer #7 · answered by Raghu 2 · 0⤊ 0⤋
9! = 9x8x7x6x5x4x3x2x1 = 362,880
2006-09-30 10:52:55 · answer #8 · answered by Kemmy 6 · 0⤊ 0⤋
your question is not specific enough to answer.
look here and ask again if you need to.
http://mathforum.org/dr.math/faq/faq.comb.perm.html
2006-09-30 07:18:11 · answer #9 · answered by horse 2 · 0⤊ 0⤋
10C10 = (10!)/((10 - 10)!)
10C10 = (10!)/(0!)
10C10 = (10!)/1
10C10 = 10!
10C10 = 3628800
2006-09-30 16:17:00 · answer #10 · answered by Sherman81 6 · 0⤊ 0⤋