Part 4/5:
- Continuing this would yield ( (n-2) ) arrangements for the third, and so on until we arrive at the base case of generating the last arrangement.
- Final Observations:
Ultimately, we discover that to sort ( n ) objects, we have:
( n ) choices for the first position,
( (n-1) ) choices for the second,
( (n-2) ), ( (n-3) ), down to 1.
We can mathematically express the total number of arrangements as:
[ n! = n \times (n-1) \times (n-2) \times \ldots \times 1 ]
This forms the basis of permutation calculations wherein the order matters.