B2. 50 Consider a random combi-number. First, we will prove that it consists of at least 50 digits. Then, we will show that there actually exists a combi-number with exactly 50 digits.

Note that if we swap the same digits everywhere in a combi-number, we still get a combi-number: for example, you can replace all threes with ones and vice versa. This way, a combi-number never needs to start with a zero.

Suppose the digit $c$ is not in the first or the last position in the combi-number. Every time $c$ appears in the combi-number, it forms a pair with both its left and right neighbor. Since $c$ must appear in 9 pairs, $c$ must therefore appear at least 5 times in the combi-number.

Suppose $c$ is the first or the last digit, but not both. Then $c$ must also appear 5 times in the combi-number, because the $c$ at the beginning (or at the end) is only in one pair. If the first and the last digit of the combi-number are the same digit $c$, then you need $c$ six times.

In total, we see that each digit from $0$ to $9$ must appear at least five times in the combi-number. Therefore, the combi-number must have at least 50 digits. It can also have exactly 50 digits, for example with the number

$$
98765432109753196307418529517394864208406283726150 .
$$

The way we created this number is as follows. We made all possible sequences of digits with a constant jump. So 98765432109 with a jump of 1, 975319 and 864208 with a jump of 2, and so on. We then concatenated these sequences in a convenient way. Of course, there are many other ways to create a 50-digit combi-number.
B2. An integer $n$ is called a combi-number if every pair of different digits from all possible digits $0 \mathrm{t} / \mathrm{m} 9$ appear next to each other at least once in the number. Thus, in a combi-number, the digits 3 and 5 appear next to each other somewhere. It does not matter whether they appear in the order 35 or 53. We agree that a combi-number does not start with the digit 0.

What is the smallest number of digits a combi-number can consist of?