Final answer:
The number of 12-bit strings containing exactly five 1's is calculated using combinatorics and the combination formula, yielding 792 distinct strings.
Step-by-step explanation:
The student's question pertains to combinatorics, a topic in mathematics that deals with counting, both in an abstract sense and in several concrete situations that involve planning and decision making. The specific question is about counting the number of 12-bit strings that contain exactly five 1's. This is equivalent to choosing 5 positions out of 12 for the 1's, and the number of ways to do this is given by the combination formula C(n, k), where n is the total number of items and k is the number of items to choose.
To calculate the number of 12-bit strings with exactly five 1's, we use the combination formula C(12, 5), which represents the number of combinations of 12 items taken 5 at a time. This can be calculated as:
C(12, 5) = \( \frac{12!}{5!(12-5)!} \) = \( \frac{12!}{5!7!} \)
After calculating the factorials and simplifying, we get:
C(12, 5) = \( \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} \) = \( \frac{95040}{120} \) = 792
Therefore, there are 792 distinct 12-bit strings that contain exactly five 1's.