anonymous
  • anonymous
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:
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
this is a permutations questions
anonymous
  • anonymous
you have 6 animals, 2 are the same and 3 are the same
anonymous
  • anonymous
they're considered distinguishableO: so not really the "same"

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anonymous
  • anonymous
the answer would be 720 if that was the case
anonymous
  • anonymous
the answer should be 60, your equation should be 6!/(3!)(2!)
anonymous
  • anonymous
the book says 120 :(
anonymous
  • anonymous
is it DISTINGUISHABLE, or INdistinguishable
anonymous
  • anonymous
distinguishiable O:
shubhamsrg
  • shubhamsrg
*
yrelhan4
  • yrelhan4
*
anonymous
  • anonymous
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)
anonymous
  • anonymous
@dydlf is it wrong ...
shubhamsrg
  • shubhamsrg
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

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