19.10 students stand in a row, and the teacher wants to give each student a hat. There are two colors of hats: red and white, and the number of hats of each color is more than 10. Requirement: in any number of consecutive adjacent students, the difference between the number of students wearing red hats and the number of students wearing white hats is at most $\mathbf{2}$. Therefore, the teacher has $\qquad$ ways to distribute the hats.
Method 1: Ordered Enumeration, Combined with Graphical Numbering Method: Moving one grid to the right represents a red hat, moving one grid up represents a white hat, a total of 10 grids to complete.
More than 2, such as points $\mathrm{a}$ and $\mathrm{b}$ cannot both exist. (Rows and columns $1 \times 4, 2 \times 5, 3 \times 6$ are not allowed, easy to overcount) Counting this way, there are 47 types below:
(To make the table fit on one page, see the next page)

According to "in any consecutive adjacent students, the difference in the number of students wearing red hats and white hats is at most 2", then among all 10 students, the number of red hats and white hats can be at most 6 red 4 white, 5 red 5 white, 4 red 6 white. Among these, 6 red 4 white and 4 red 6 white are symmetrical, with the same number of types.
(1) 6 red divided into three groups, red red, red red, red red
4-2=2, between red red and red red there must be two whites, 1 type
$\checkmark \checkmark, \times \times, \vee \checkmark, \times$ subtotal, 6 red divided into three groups total 1 type;
(2) 6 red divided into four groups, red red, red red, red, red
(1) red red, red red, red, red Between red red and red red there must be two whites, 1 type
$\sqrt{ } \sqrt{ } \times \times, \vee \sqrt{ }$, (2) red, red, red red, red red
(3) red red, red, symmetry, 1 type;
5-2=3, in these two intervals, one must be 1 white, the other must be two whites, 2 types
(4) red, red red, red, red red
(5) same as (3), 2 and $\qquad$
$\qquad$ 6-2=4, the ends cannot possibly be white, 3 types:
(6) red, red red, red red,

Between red red and red red there must be two whites, 1 type

Subtotal, 6 red divided into five groups total $2+3=5$ types; Therefore, 6 red 4 white total $1+10+5=16$ types;
(2) 4 red 6 white same as 6 red 4 white, total 16 types
(3) 5 red 5 white (1) 5 red divided into three groups, red red, red red, red
$\qquad$
$\times, \vee \checkmark, \times \times$, (2) red, red red, red red
Same as (1), by symmetry, 5 types;
$\qquad$
2) 5 red divided into four groups, red red, red, red, red (1) red red, red, red, red
(1) red red, red, red, red $1+2=3,2+2=4$, the three intervals marked with lines are 3 to 4 whites, 9 types $\qquad$
$\qquad$
$\qquad$
$\qquad$
(2) red,
(3)
(3)
(3) same as (1), by symmetry, 9
(3) red, $1+2=$ 11 types
$1+2=3$, 11 types: $\sqrt{ }, \quad \times$. $\begin{array}{l} 11 \\ \checkmark \text { types: } \\ \checkmark \\ \checkmark \times\end{array}$ $\qquad$
$\qquad$
$\qquad$
$\qquad$
(4) red,
same as (3)
small same as (3)
11 types;
$\qquad$
$\qquad$
$\qquad$
$\qquad$
$\qquad$
$\qquad$
(3) 5 red $\square$
$\square$
$\qquad$
$\qquad$
$\qquad$
$\qquad$
$\qquad$
In summary, a total of 16+16+62=94 types