Final answer:
There are 84 seven-digit numbers that have their digits arranged in strictly decreasing order. This is determined by the combination formula C(9, 6), which accounts for the selection of 6 digits excluding 0 for the first position from the numbers 1 through 9.
Step-by-step explanation:
To find out how many 7-digit numbers have their digits arranged in strictly decreasing order, we must select 7 different digits from 0 to 9 and arrange them in descending order. Since the order is fixed once the digits are chosen, this is a combination problem rather than a permutation problem. We have 10 digits to choose from (0, 1, 2, 3, 4, 5, 6, 7, 8, 9), and we need to select 7 of them, with the restriction that 0 cannot be the leftmost digit.
The number of ways to choose 7 decreasing digits from 10 without having 0 as the first digit is the same as choosing 6 digits from the 9 remaining (since one of the chosen digits will be the leftmost digit and cannot be 0). Therefore, we use the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of items to choose from, and k is the number of items to pick.
Here, we get C(9, 6) because we are choosing 6 digits from 9 options (1 through 9). This results in:
C(9, 6) = 9! / (6! * (9-6)!) = 84
So, there are 84 such 7-digit numbers whose digits are in strictly decreasing order.