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DLS

  • 2 years ago

Simplify..

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  1. DLS
    • 2 years ago
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    \[C(40,31) + summation(from j=0 \to j=10) C(40+j,10+j)\]

  2. DLS
    • 2 years ago
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    someone please put it nicely in an equation form!

  3. DLS
    • 2 years ago
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    |dw:1351702148391:dw|

  4. sauravshakya
    • 2 years ago
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    |dw:1351703115405:dw|

  5. sauravshakya
    • 2 years ago
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    |dw:1351703210822:dw|

  6. sauravshakya
    • 2 years ago
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    |dw:1351703313107:dw|

  7. sauravshakya
    • 2 years ago
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    Am I correct @experimentX ???

  8. experimentX
    • 2 years ago
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    |dw:1351703384969:dw|

  9. sauravshakya
    • 2 years ago
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    Oh yes! that would decrease one step.

  10. experimentX
    • 2 years ago
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    keep using this, probably you would end up with \[ \binom{51}{31}\]

  11. experimentX
    • 2 years ago
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    something like that ... not so sure.

  12. sauravshakya
    • 2 years ago
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    |dw:1351703789879:dw|

  13. DLS
    • 2 years ago
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    howdid u solve C(40,31) & C(40,10) thats the main thing !

  14. DLS
    • 2 years ago
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    just explain this part

  15. DLS
    • 2 years ago
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    and its 51 C 20! not 21..

  16. experimentX
    • 2 years ago
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    C(40,10) = C(40,30)

  17. sauravshakya
    • 2 years ago
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    |dw:1351703959025:dw|

  18. DLS
    • 2 years ago
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    ??

  19. sauravshakya
    • 2 years ago
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    |dw:1351704198408:dw|

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