nintso wrote:
Hello,
I have a question to Bunuel. the formula you are using is Combination's formula as it is 6!/(6-3)!*3!, and since Alice has to choose 3 Ss and 3Es, where order does not matter, it has to be Combination. but you are saying that "Total # of routs to the school is # of permutation of SSSEEE, which is 6!/(3!3!)=20 (# of permutations of 6 letters out of which 3 S's and 3 E's are identical)"[u][/u]. can you please explain whether it should be a combination or permutation formula?
I am having hard time understanding this problem. Manhattan guide explains it with anagram grid, but I am not grasping that model. I tried to solve it using Slot Method, but could not do it. I can't really get it with combination/permutation formulas either. can you please explain it in an easier way with more details?
I have two keywords for Permutations and combinations -
Arrangement and Selection.
SELECTION means -
CHOOSING 1, more or nothing. (
Combinations)
ARRANGEMENT means -
RE-ORGANIZING or ORDER(
Permutations)
In this given problem, Alice should definitely take 3 souths and 3 Easts to reach her school. But in any 'ORDER' of her choice....
==> I need to use Permutations and not combinations formula as I hear the word 'ORDER'
so calculating the total number of permutations = \(6!/3!3!\)= 20
If Alice needs to take 2 Souths first, then the remaining 4 steps - 1 South and 3 Easts can be
re-arranged (user Permutations) in = \(4!/3!\)= 4
Probability = \(4/20 = 1/5\)