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anonymous

  • one year ago

math

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  1. hartnn
    • one year ago
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    what have you tried?? (a) If first 3 MUST be answered, then how many are remaining, and how many out of them are to be answered?

  2. anonymous
    • one year ago
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    oh so its 7c5

  3. hartnn
    • one year ago
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    thats right :)

  4. hartnn
    • one year ago
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    (b) Atleast 4 out of 1st 5, so 2 cases : 4 out of first 4 and another 4 out of another 5 OR 5 out of 1st 5 and another 3 out of remaining 5

  5. hartnn
    • one year ago
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    ** 4 out of first 5 and another 4 out of another 5 OR 5 out of 1st 5 and another 3 out of remaining 5

  6. anonymous
    • one year ago
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    2c1?

  7. hartnn
    • one year ago
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    4 out of first 5 = 4C5 4 out of another 5 = 4C5 "4 out of first 5 and another 4 out of another 5" = 4C5 + 4C5 do it similarly for "5 out of 1st 5 and another 3 out of remaining 5"

  8. hartnn
    • one year ago
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    sorry *** "4 out of first 5 and another 4 out of another 5" = 4C5 * 4C5

  9. hartnn
    • one year ago
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    woah! 5C4, not 4C5 :P

  10. anonymous
    • one year ago
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    Omg. 2c1 so far from 5c4. Darn.

  11. anonymous
    • one year ago
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    Thank you so much @hartnn

  12. hartnn
    • one year ago
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    so you got it? and welcome ^_^

  13. anonymous
    • one year ago
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    yes yes =)

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