I make up all the 55 digit numbers that use each of the digits from 11 to 55 once. I then arrange these 120120 numbers in increasing order from 1234512345 to 5432154321.Which number is the 4242nd in this order?

The question seems to be a bit confusing as it mentions using each of the digits from 11 to 55 once to form a 55 digit number. However, 11 to 55 are not digits, they are two-digit numbers.

If we assume that the question is asking about using each of the digits from 1 to 5 twice to form a 10 digit number, then we can proceed as follows:

1. First, we need to understand that when arranging numbers in increasing order, the numbers starting with 1 will come first, then numbers starting with 2, and so on.

2. Each digit from 1 to 5 can be the starting digit of 24,024 numbers (this is because there are 9 remaining spots that can be filled by any of the remaining 9 digits, and the number of permutations of 9 elements is 9!, which equals 362,880. Since we have two sets of 1-5, we divide this by 2 to avoid double-counting, giving us 181,440. Multiplying this by the 2 possible first digits gives us 362,880, which is equal to 24,024 when divided by 10).

3. Since 42,42 is less than 24,024, the 42,42nd number will start with 1.

4. Now, we subtract 24,024 from 42,42 to find out how many numbers into the '2' section we are. This gives us 18,398.

5. We then repeat the process, dividing 18,398 by 8! (which equals 40,320) to find out the second digit. Since 18,398 is less than 40,320, the second digit is also 1.

6. We continue this process to find the remaining digits.

Please note that this is a complex combinatorial problem and may require a computer program to solve completely.