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:
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this is a permutations questions
you have 6 animals, 2 are the same and 3 are the same
they're considered distinguishableO: so not really the "same"
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the answer would be 720 if that was the case
the answer should be 60, your equation should be 6!/(3!)(2!)
the book says 120 :(
is it DISTINGUISHABLE, or INdistinguishable
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)
is it wrong ...
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