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Ishaan94

  • 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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  1. Ishaan94
    • 3 years ago
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    @satellite73 @eliassaab

  2. Ishaan94
    • 3 years ago
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    @experimentX :D

  3. anonymous
    • 3 years ago
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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

  4. anonymous
    • 3 years ago
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    without thinking too hard i get an answer that could easily be wrong, namely \[9+8+7+6+5+4+3+2\]

  5. anonymous
    • 3 years ago
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    oops +1 at the end because \(a_1\) could be 9th leaving one possibility for \(a_2\)

  6. anonymous
    • 3 years ago
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    oh crap that is all wrong

  7. anonymous
    • 3 years ago
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    but idea is probably right. if \(a_1\) is first there are 9! arrangements

  8. experimentX
    • 3 years ago
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    I think it could be 1(9+8+7+6+5+4+3+2+1)8!

  9. Ishaan94
    • 3 years ago
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    yeah

  10. anonymous
    • 3 years ago
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    if \(a_1\) is second there are 8 times 8! arrangements

  11. experimentX
    • 3 years ago
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    |dw:1340650357123:dw|

  12. anonymous
    • 3 years ago
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    so yeah, experimentX has it

  13. Ishaan94
    • 3 years ago
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    experimentx's makes sense. and so does satellite's idea. Medal each other.

  14. Ishaan94
    • 3 years ago
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    experiemntx's answer*

  15. Ishaan94
    • 3 years ago
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    thank you. :-)

  16. anonymous
    • 3 years ago
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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

  17. experimentX
    • 3 years ago
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    well that solves it ;)

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