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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
without thinking too hard i get an answer that could easily be wrong, namely \[9+8+7+6+5+4+3+2\]
oops +1 at the end because \(a_1\) could be 9th leaving one possibility for \(a_2\)
oh crap that is all wrong
but idea is probably right. if \(a_1\) is first there are 9! arrangements
I think it could be 1(9+8+7+6+5+4+3+2+1)8!
if \(a_1\) is second there are 8 times 8! arrangements
so yeah, experimentX has it
experimentx's makes sense. and so does satellite's idea. Medal each other.
thank you. :-)
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
well that solves it ;)