True.
If you have [tex] n [/tex] objects, the most common and important combinatorics elements you have are:
- Permutations: you use every element, and the point is: "how do you order them?".
- Dispositions: you choose a subset of [tex] k<n [/tex] elements, and consider a specific position for each element in the subset. So, you're asking "how many ordered subsets of [tex] k [/tex] elements can I get from my [tex] n [/tex]-elements set?
- Combinations: you choose a subset of [tex] k<n [/tex] elements. So, you're asking "how many subsets of [tex] k [/tex] elements can I extract from my [tex] n [/tex]-elements set?
Here are some examples: if you start from the set [tex] \{1,2,3\} [/tex], all the possible permutations are
[tex] \{1,2,3\},\ \{1,3,2\},\ \{2,1,3\}, \{2,3,1\}, \{3,1,2\}, \{3,2,1\} [/tex]
All the possible dispositions, choosing the subset cardinality [tex] k=2[/tex], are
[tex] \{1,2\},\ \{2,1\},\ \{1,3\},\ \{3,1\},\ \{2,3\},\ \{3,2\} [/tex]
So, as you can see, the order matters, because [tex] \{1,2\} [/tex] and [tex] \{2,1\} [/tex] are not the same element.
Finally, the combinations (still with [tex] k=2 [/tex]) are
[tex]\{1,2\},\ \{1,3\},\ \{2,3\}[/tex]
Because now the order doesn't matter, and thus [tex] \{1,2\} [/tex] and [tex] \{2,1\} [/tex] are seen as the same element.
