12. Six positive integers $a, b, c, d, e, f$ are arranged in alphabetical order to form a sequence, where $a=1$. If a certain positive integer is greater than 1, then the number that is one less than this positive integer must appear to its left. For example, if $d>1$, then one of $a, b, c$ must be $d-1$. Examples: $1,1,2,1,3,2$ satisfies the condition; $1,2,3,1,4,1$ satisfies the condition; $1,2,2,4,3,2$ does not satisfy the condition. The number of different permutations that satisfy the condition is $\qquad$.
【Answer】 203
【Solution】 We use $b_{n}$ to denote the number of sequences of $n$ natural numbers that satisfy the requirements of the problem. Clearly, $b_{1}=1$. Next, we calculate $b_{2}$, which is a sequence of two positive integers. Since $a=1$, then $b=1$ or 2. If $\left\{\begin{array}{l}a=1 \\ b=1\end{array}\right.$, then the largest number in the sequence is 1, and there is 1 such sequence; if $\left\{\begin{array}{l}a=1 \\ b=2\end{array}\right.$, then the largest number in the sequence is 2, and there is 1 such sequence. We use the following notation for counting:
$b_{2}=$\begin{tabular}{ccc}
1 & + & 1 \\
$\uparrow$ & & $\uparrow$ \\
the largest number in the sequence is 1 & & the largest number in the sequence is 2 \\
\hline
\end{tabular}$;$

Next, we calculate $b_{3}$, which is a sequence of three positive integers. (1) To make the largest number in the sequence 1, the preceding numbers can only be 1, so there is 1 such sequence. (2) To make the largest number in the sequence 2, there are two possibilities: the preceding numbers are all 1, and the last number is 2, which gives 1 such sequence, this 1 is the first number in the previous $b_{2}$; the preceding numbers have already reached 2, so the last number can be 1 or 2, giving $1 \times 2$ sequences. The 1 is the second number in the previous $b_{2}$, and 2 indicates that the last number has two ways to be written; (3) To make the largest number in the sequence 3, the preceding two numbers must form a sequence where the largest number is 2, which gives 1 such sequence, this 1 is the second number in the previous $b_{2}$. We use the following notation for counting:
\begin{tabular}{|c|c|c|c|c|}
\hline 1 & \multirow[t]{2}{*}{+} & $1+1 \times 2$ & \multirow[t]{2}{*}{+} & 1 \\
\hline$\uparrow$ & & $\uparrow$ & & $\uparrow$ \\
\hline the largest number in the sequence is 1 & & the largest number in the sequence is 2 & & the largest number in the sequence is 3 \\
\hline 1 & + & 3 & + & 1 \\
\hline$\uparrow$ & & $\uparrow$ & & $\uparrow$ \\
\hline the largest number in the sequence is 1 & & the largest number in the sequence is 2 & & the largest number in the sequence is 3 \\
\hline
\end{tabular}

Using the same reasoning, we obtain the following results:
$$
\begin{aligned}
b_{5} & =1+(1+7 \times 2)+(7+6 \times 3)+(6+1 \times 4)+1 \\
& =1+15+25+10+1 \\
b_{6} & =1+(1+15 \times 2)+(15+25 \times 3)+(25+10 \times 4)+(10+1 \times 5)+1 \\
& =1+31+90+65+15+1=203
\end{aligned}
$$