Final answer:
The total number of bit strings of length 12 containing exactly four 1s is 495. This is obtained by calculating the combinations of 4 positions out of 12 using the formula C(12, 4), which simplifies to 12! divided by 4! times 8!, yielding 495.
Step-by-step explanation:
To calculate the total number of bit strings of length 12 containing exactly four 1s, we must understand the concept of combinations in mathematics. A bit string is a sequence of bits (0s and 1s), and in this case, we want a string 12 bits long with exactly four 1s and the rest being 0s. Since the order in which the 1s are placed does not matter, only which positions they occupy, we use combinations to find the answer.
The number of ways to choose 4 positions out of 12 for the 1s is determined by the combination formula, which is typically written as C(n, k) or sometimes nCk, where n is the total number of items to choose from, and k is the number of items to choose. The formula for this is C(n, k) = n! / (k!(n - k)!), where '!' represents the factorial of a number. The factorial of a number, such as 4!, is the product of all positive integers up to that number, so 4! = 4×3×2×1.
Using this formula, we calculate C(12, 4) = 12! / (4!(12 - 4)!) = (12×11×10×9×8×7×6×5×4×3×2×1) / ((4×3×2×1)(8×7×6×5×4×3×2×1)), which simplifies after canceling out the common terms to 12×11×10×9 / 4×3×2×1. This gives us a total of 495 bit strings of length 12 containing exactly four 1s.