luisnavarro wrote:
Hi Bunuel,
Sorry, I am a little confused yet, 5! gives different arrangements without 0 {1, 2, 3, 4, 5}, and in your explanation you said:
5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2, 3, 0, 4, 5}, ... Which is not what the question is asking.
But I was reffering to 5! without cero... I am confused, could you help me again?
Thanks a lot
Regards
Luis Navarro
Looking for 700
Yes, 5! gives different arrangements of {1, 2, 3, 4, 5}. 0 in my previous post was a typo.
We need SUBSETS of {0, 1, 2, 3, 4, 5} without 0, not arrangements of {1, 2, 3, 4, 5}. Subsets of {0, 1, 2, 3, 4, 5} without 0 are:
{empty};
{1};
...
{5}
{1, 2}
{1, 3}
...
...
{1, 2, 3, 4, 5}
Notice that arrangements of {1, 2, 3, 4, 5} give the same 5-element sets but arranged in different ways, while subsets give an empty set, 1-element sets, 2-elements set, ..., 5-element sets.
Hope it's clear.
Thanks, it is more clear know, I only have a little more doubt...
What I inffer for this is that subsets orders by itself "automaticly", so then {1, 2, 3, 4, 5} only consider 1 posible combination, instead of mixing the order in the positions for instance: {3, 1, 5, 4, 2}, {2, 5, 3, 1, 4}... is my conclusion wrong or correct?
Thanks a lot.
Regards.