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dydlf

heeelp! in how many ways can 2 elephants, 1 monkey and 3 ants sit in a row if no 2 animals of the same kind are together? assume 2 animals of the same kind are distinguishable. answer is said to be 120. O:

  • one year ago
  • one year ago

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  1. cammyabbo
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    this is a permutations questions

    • one year ago
  2. cammyabbo
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    you have 6 animals, 2 are the same and 3 are the same

    • one year ago
  3. dydlf
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    they're considered distinguishableO: so not really the "same"

    • one year ago
  4. cammyabbo
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    the answer would be 720 if that was the case

    • one year ago
  5. cammyabbo
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    the answer should be 60, your equation should be 6!/(3!)(2!)

    • one year ago
  6. dydlf
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    the book says 120 :(

    • one year ago
  7. cammyabbo
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    is it DISTINGUISHABLE, or INdistinguishable

    • one year ago
  8. dydlf
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    distinguishiable O:

    • one year ago
  9. shubhamsrg
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    *

    • one year ago
  10. yrelhan4
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    *

    • one year ago
  11. matricked
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    first arrange the elephants and the ants in the way such that the elephants are in betweeb the ants as such A1 E1 A2 E2 A3 which can be done in 3!*2! ways now the monkey can be paired with any of the five animals in 5*2! ways =10 e.g A1 E1M1 A2 E2 A3 & A1 M1E1 A2 E2 A3 (here paired with E1) another one paired with A3 A1 E1 A2 E2 A3 M1 & A1 E1 A2 E2 M1A3 in this way the number of ways is 3!*2!*(10)=120 (as no two animals of same kind are together)

    • one year ago
  12. matricked
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    @dydlf is it wrong ...

    • one year ago
  13. shubhamsrg
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    possible no. of arrangements of 3 ants is : 1) a_a_a_ -> 3! * 3! 2) a__a_a ->(3! * 3!) - ( 3! * 2!) 3) a_a__a -> (3! * 3!) - (31 * 2!) 4) _a_a_a -> 3! *3! so final answer= 36 +24 +36+ 24 = 120

    • one year ago
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