Final answer:
Considering the prime factorization of 3072 and the constraints of using only 1, 2, 3, 4 for the digits, the number of eight-digit combinations such that their product is 3072 is found to be 8. Thus, k = 8.
Step-by-step explanation:
The student asked about the number of ways an eight-digit number can be formed from the digits 1, 2, 3, 4 such that the product of all digits is always 3072. To find k, we must first decompose 3072 into prime factors to see how many times each number (1-4) will be used in the eight-digit number. The prime factorization of 3072 is 210 × 3, which means in a combination of eight digits we'll have ten 2's and one 3. We know 1 does not affect the product, so any number of 1's can be included so long as the count of digits totals eight. However, since we're limited to using the digits 1-4, the number 3 can only appear once, and the rest must be 2's and 1's to maintain the product of 3072. The problem then simplifies to finding the number of ways to arrange seven 2's and one 3, which is simply 8! (the number of ways to arrange 8 distinct items) divided by 7! (the number of ways to arrange the seven identical items, which is 2), calculating as 8 combination ways. Therefore, k is 8.