Number of arrangements of the set $\{a_1,a_2, \ldots, a_{10}\}$, so that $a_1$ is always ranked above $a_2$.

Question

anonymous · Answer

@satellite73 @eliassaab

anonymous · Answer

@experimentX :D

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

oh crap that is all wrong

anonymous · Answer

but idea is probably right.
if $a_1$ is first there are 9! arrangements

experimentx · Answer

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

anonymous · Answer

yeah

anonymous · Answer

if $a_1$ is second there are 8 times 8! arrangements

experimentx · Answer

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