Ishaan94
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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Ishaan94
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@satellite73 @eliassaab
Ishaan94
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@experimentX :D
anonymous
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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
anonymous
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without thinking too hard i get an answer that could easily be wrong, namely
\[9+8+7+6+5+4+3+2\]
anonymous
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oops +1 at the end because \(a_1\) could be 9th leaving one possibility for \(a_2\)
anonymous
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oh crap that is all wrong
anonymous
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but idea is probably right.
if \(a_1\) is first there are 9! arrangements
experimentX
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I think it could be 1(9+8+7+6+5+4+3+2+1)8!
Ishaan94
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yeah
anonymous
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if \(a_1\) is second there are 8 times 8! arrangements
experimentX
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|dw:1340650357123:dw|
anonymous
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so yeah, experimentX has it
Ishaan94
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experimentx's makes sense. and so does satellite's idea.
Medal each other.
Ishaan94
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experiemntx's answer*
Ishaan94
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thank you. :-)
anonymous
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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
experimentX
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well that solves it ;)