Moving horizontally and vertically from point to point along the lines in the diagram below, how many routes are there from point $A$ to point $B$ which consist of six horizontal moves and six vertical moves?

[asy]
for(int i=0; i<=6;++i)
{
draw((i,i)--(6,i),black);
draw((i,i)--(i,0),black);
for(int a=i; a<=6;++a)
{
dot((a,i));
}
}
label("A",(0,0),W);
label("B",(6,6),N);

[/asy]
1. To solve the problem of finding the number of routes from point \( A \) to point \( B \) consisting of six horizontal moves and six vertical moves, we can use combinatorial methods. Specifically, we need to determine the number of ways to arrange 6 horizontal moves (H) and 6 vertical moves (V) in a sequence of 12 moves.

2. The total number of moves is 12, and we need to choose 6 positions out of these 12 for the horizontal moves (the remaining 6 will automatically be vertical moves). This is a classic combination problem, which can be solved using the binomial coefficient:
   \[
   \binom{12}{6}
   \]

3. The binomial coefficient \(\binom{n}{k}\) is calculated as:
   \[
   \binom{n}{k} = \frac{n!}{k!(n-k)!}
   \]
   where \( n = 12 \) and \( k = 6 \).

4. Substituting the values, we get:
   \[
   \binom{12}{6} = \frac{12!}{6!6!}
   \]

5. Calculating the factorials:
   \[
   12! = 479001600
   \]
   \[
   6! = 720
   \]

6. Substituting these values back into the binomial coefficient formula:
   \[
   \binom{12}{6} = \frac{479001600}{720 \times 720} = \frac{479001600}{518400} = 924
   \]

7. Therefore, the number of routes from point \( A \) to point \( B \) is:
   \[
   \boxed{924}
   \]