For question 1:

Here's how to find the number of positive five-digit integers containing the digit grouping "24" at least once:

Total Possible Numbers:

There are 9 choices (excluding 0) for each digit in a five-digit number (1, 2, 3, 4, 5, 6, 7, 8, 9).

So, the total number of possible five-digit positive integers is 9 * 9 * 9 * 9 * 9 = 59,049.

Unwanted Scenarios (Numbers without "24"):

No "24" group: There are 8 choices for the digit in the place where "24" could be (all digits except 2 and 4). For the remaining 4 digits, there are 9 choices each. So, there are 8 * 9 * 9 * 9 = 5832 cases where "24" doesn't appear in any of the five digits.

"24" only appears once, but not next to each other:

Choose the digit that will be between "2" and "4": There are 7 choices (all digits except 0, 2, and 4).

Choose the positions for "2" and "4" (not next to each other): There are 4 choices for the first position (remaining slots after fixing one digit). There are 3 choices for the second position (remaining slots after fixing the first and the middle digit). So, there are 4 * 3 = 12 ways to arrange "2" and "4" with another digit in between.

Choose the remaining 2 digits: There are 9 choices each.

Total unwanted cases with "24" not next to each other: 7 * 12 * 9 * 9 = 5832.

Total Numbers with "24" at least once:

We want to find the number with "24" at least once. So, subtract the unwanted cases from the total:

Total numbers - Unwanted cases = Numbers with "24"

59,049 - (5832 + 5832) = 47,385

Therefore, there are 47,385 positive five-digit integers containing the digit grouping "24" at least once.