Question

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:

Answer 1

this is a permutations questions

Answer 2

you have 6 animals, 2 are the same and 3 are the same

Answer 3

they're considered distinguishableO: so not really the "same"

Answer 4

the answer would be 720 if that was the case

Answer 5

the answer should be 60, your equation should be 6!/(3!)(2!)

Answer 6

the book says 120 :(

Answer 7

is it DISTINGUISHABLE, or INdistinguishable

Answer 8

distinguishiable O:

Answer 9

*

Answer 10

*

Answer 11

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)

Answer 12

@dydlf

is it wrong ...

Answer 13

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