anonymous
  • anonymous
Number of arrangements of the set \(\{a_1,a_2, \ldots, a_{10}\}\), so that \(a_1\) is always ranked above \(a_2\).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@satellite73 @eliassaab
anonymous
  • anonymous
@experimentX :D
anonymous
  • anonymous
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

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More answers

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

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