Ishaan94 3 years ago Number of arrangements of the set \(\{a_1,a_2, \ldots, a_{10}\}\), so that \(a_1\) is always ranked above \(a_2\).

1. Ishaan94

@satellite73 @eliassaab

2. Ishaan94

@experimentX :D

3. satellite73

dunno i hate counting but if \(a_1\) is first then there are 9 choices for \(a_2\) if \(a_1\) is second there are 8, if \(a_1\) is third there are 7, so i guess we can use that

4. satellite73

without thinking too hard i get an answer that could easily be wrong, namely \[9+8+7+6+5+4+3+2\]

5. satellite73

oops +1 at the end because \(a_1\) could be 9th leaving one possibility for \(a_2\)

6. satellite73

oh crap that is all wrong

7. satellite73

but idea is probably right. if \(a_1\) is first there are 9! arrangements

8. experimentX

I think it could be 1(9+8+7+6+5+4+3+2+1)8!

9. Ishaan94

yeah

10. satellite73

if \(a_1\) is second there are 8 times 8! arrangements

11. experimentX

|dw:1340650357123:dw|

12. satellite73

so yeah, experimentX has it

13. Ishaan94

experimentx's makes sense. and so does satellite's idea. Medal each other.

14. Ishaan94

15. Ishaan94

thank you. :-)

16. satellite73

that answer makes sense to me. i was totally wrong the first time, and didn't know that formula but now that i see it it is clear

17. experimentX

well that solves it ;)