the digits 6 7 and 8 exactly once in each number is this permutation or combination?

Answer 1

let us say a,b,c are three numbers (or any objects)...

possible arrangements:

abc acb bca bac cab cba (6); here the same elements are repeated....only order is different

suppose we select only 2 of them at a time, then:

ab ac bc ba ca cb (6)

in this, ab & ba have the same elements... so this is permutation

if we eliminate the same element combinations, we get

ab ac bc .... this is combination

permutation considers the order of elements also (ab is different from ba)

combination does not consider the order of elements

so , permutation will be always equal or greater than combination...

if we select numbers such that 6,7,8 appear exactly once, then we are not considering the order of them...(678 is different from 768 etc).... so this is permutation

Answer 2

you weren't specific enough. for example, are you talking about three cipher numbers? i.e. 99
permutation is bijective mapping.

if S is bijective mapping (bijection), then you can write
S: {6,7,8} ->{6,7,8} ...
then you have for all x,y e {6,7,8} S(x)=X(y) => x=y ... and, for any y e {6,7,8} there exists x e {6,7,8}, for which S(x)=y.

when defined like that, S is a permutation, and when you follow those rules, you get that any element of mapping S is actually an element of permutation. ... and so on