1. ## Permatations

OK... I am in the procsess of building a spreadsheet to help me with route planning but I think I may have stumbled on a problem that someone might be able to help me with.

To put it simply, I have 14 objects and need to list the different combinations that they can be ordered into.

e.g 1,2,3,4,5,6,7,8,910,11,12,13,14
or
1,3,2,4,5,6,7,8,9,10,11,12,13,14

Am I right in saying that there are over eighty seven billion results / combinations or am I working this out wrong?

2. ## Re: Permatations

There are 14! ways to arrange the numbers 1-14

This turns out to be about $8.72 \cdot 10^{10}$

3. ## Re: Permatations

$14!=87178291200$

4. ## Re: Permatations

OK.

Well is there any way to find out the number of results under certain rules?

1 has to be before 2
3 has to be before 4
5 has to be before 6
7 has to be before 8
9 has to be before 10
11 has to be before 12
13 has to be before 14

By before I mean to the left of. so it can't start with 2,1 etc.

Is there anyway I can find out how many that leaves me with?

5. ## Re: Permatations

Originally Posted by thegooser123
Well is there any way to find out the number of results under certain rules?
1 has to be before 2
3 has to be before 4
5 has to be before 6
7 has to be before 8
9 has to be before 10
11 has to be before 12
13 has to be before 14
Is there anyway I can find out how many that leaves me with?
As I understand it, we could have: $1,13,2,11,5,7,3,14,6,9,8,12,4,10.$
Is that correct.
If so, there are $\frac{14!}{2^7}=681080400$ possible such strings.

6. ## Re: Permatations

Could you tell me what the results would be with the same criteria but with numbers 1 t0 8 and also numbers 1 to 6?

7. ## Re: Permatations

for 8 it would be 2520,and for 6 it would be 90.

9. ## Re: Permatations

Originally Posted by thegooser123
Could you tell me what the results would be with the same criteria but with numbers 1 t0 8 and also numbers 1 to 6?
If we go with these rules
1 has to be before 2
3 has to be before 4
5 has to be before 6
7 has to be before 8
then $\frac{8!}{2^4}$

In general for $\{1,2,\dots,2n\}$, $\frac{(2n)!}{2^n}$.

10. ## Re: Permatations

Originally Posted by Plato
If we go with these rules
1 has to be before 2
3 has to be before 4
5 has to be before 6
7 has to be before 8
then $\frac{8!}{2^4}$

In general for $\{1,2,\dots,2n\}$, $\frac{(2n)!}{2^n}$.

Plato, could you maybe explain how you came to that formula?

Thanks!

11. ## Re: Permatations

Originally Posted by Also sprach Zarathustra
Plato, could you maybe explain how you came to that formula?
If $n=5$ then there are $10!$ ways to arrange the $2(5)$ numbers.
If half of them the 1 comes before the 2. Also in half the 3 comes before the 4. Just continue that pattern.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}