Respuesta :
Answer:
The answer to this question can be defined as follows:
Explanation:
Its Permute-with-all method, which doesn't result in a consistent randomized permutation. It takes into account this same permutation, which occurs while n=3. There's many 3 of each other, when the random calls, with each one of three different values returned and so, the value is= 27. Allow-with-all trying to call possible outcomes as of 3! = 6
Permutations, when a random initial permutation has been made, there will now be any possible combination 1/6 times, that is an integer number m times, where each permutation will have to occur m/27= 1/6. this condition is not fulfilled by the Integer m.
Yes, if you've got the permutation of < 1,2,3 > as well as how to find out design, in which often get the following with permute-with-all chances, which can be defined as follows:
[tex]\bold{PERMUTATION \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ PROBABILITY}[/tex]
[tex]\bold{<1,2,3> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4/27= 0.14 }\\\bold{<1,3,2> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5/27=0.18}\\[/tex]
[tex]\bold{<2,1,3> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5/27=0.18}\\\bold{<2,3,1>\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5/27=0.18}[/tex]
[tex]\bold{<3,2,1> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4/27=0.14}[/tex]
Although these ADD to 1 none are equal to 1/6.