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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 @eliassaab

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dunno 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
 3 years ago
Best ResponseYou've already chosen the best response.0without thinking too hard i get an answer that could easily be wrong, namely \[9+8+7+6+5+4+3+2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oops +1 at the end because \(a_1\) could be 9th leaving one possibility for \(a_2\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh crap that is all wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but idea is probably right. if \(a_1\) is first there are 9! arrangements

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1I think it could be 1(9+8+7+6+5+4+3+2+1)8!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if \(a_1\) is second there are 8 times 8! arrangements

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1340650357123:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so yeah, experimentX has it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0experimentx's makes sense. and so does satellite's idea. Medal each other.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0experiemntx's answer*

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that 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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1well that solves it ;)
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