Explanation:
Selection sort is a simple comparison-based sorting algorithm. It works by dividing the input array into two parts: the sorted part and the unsorted part. In each iteration, the smallest element from the unsorted part is selected and swapped with the element at the beginning of the unsorted part. This process is repeated until the entire array is sorted.

To find the tightest upper bound that represents the number of swaps required to sort n numbers using selection sort, we need to analyze the algorithm's behavior.

Analysis:
- In the first iteration, the smallest element is selected and swapped with the first element. This requires 1 swap.
- In the second iteration, the second smallest element is selected and swapped with the second element. This requires 1 swap.
- In the third iteration, the third smallest element is selected and swapped with the third element. This requires 1 swap.
- ...

In general, in the i-th iteration, the i-th smallest element is selected and swapped with the i-th element. This requires 1 swap.

Therefore, the total number of swaps required to sort n numbers using selection sort can be calculated as follows:

Total number of swaps = 1 + 1 + 1 + ... + 1 (n times)

This can be simplified to:

Total number of swaps = n

Tightest Upper Bound:
The tightest upper bound represents an algorithm's worst-case time complexity. In the case of selection sort, the worst-case scenario occurs when the input array is in reverse order. In this case, every element needs to be swapped with every other element.

Therefore, the tightest upper bound that represents the number of swaps required to sort n numbers using selection sort is O(n).

Conclusion:
Option 'B' (O(n)) is the correct answer as it represents the tightest upper bound for the number of swaps required to sort n numbers using selection sort.