Celine traces paths on the grid below starting at point $X$ and ending at point $Y$. Each path must follow the lines connecting the dots and only ever move horizontally to the right or vertically down. It may be useful to know that there are a total of 924 such paths. Consider the 7 points labelled $A, B, C, D, E, F$, and $G$. List these points in decreasing order of the number of paths passing through that point. For example, $B$ is on exactly one path, so $B$ should be the last point in your list.

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Denote by $f(A)$ the number of paths through $A, f(B)$ the number of paths through $B$, and so on.

$f(A)$ : Every path must pass through either point $A$ or the point directly below $X$.

No path can pass through both of these points since this would require that the path either move up or to the left.

By symmetry, the same number of paths must pass through $A$ and the point directly below $X$, so $f(A)=\frac{924}{2}=462$.

$\underline{f(B)}$ : As noted in the problem statement, $f(B)=1$ since there is only one path through $B$.

$\underline{f(C)}$ : We can compute $f(C)$ by multiplying the number of paths from $X$ to $C$ by the number of paths from $C$ to $Y$.

Every path from $X$ to $C$ takes a total of 5 steps, 3 to the right and 2 down.

Thus, we can think of a path from $X$ to $C$ as a sequence of three $r$ 's and two $d$ 's. There are ten such sequences:

$r r r d d, r r d r d, r r d d r, r d r r d, r d r d r, r d d r r, d r r r d, d r r d r, d r d r r, d d r r r$.

Another way to think about counting these sequences is to count the number of ways to place the two d's in any of the five positions. This completely determines the sequence since the other three positions must be occupied by $r$ 's.

There are five positions for the first $d$, then four positions for the second $d$ for a total of $5 \times 4=20$. However, we will have counted each sequence twice since the order of the $d$ 's does not matter.

Therefore, there are a total of $\frac{20}{2}=10$ sequences of two $d$ 's and three $r$ 's, which means there are a total of 10 paths from $X$ to $C$.

To count the number of paths from $C$ to $Y$, we can similarly count the number of sequences of three $r$ 's and four d's.

This can be done by counting the number of ways to place the three $r$ 's in any of the seven possible positions.

Once again, there are 7 positions for the first $r$, then 6 remaining for the second $r$, and 5 for the third $r$ for a total of $7 \times 6 \times 5=210$ sequences.

We have again overcounted, so we need to determine by how much.

If we imagine that the three $r$ 's have different "colours" (say red, blue, and green, for example), then there are 6 possible orders in which the $r$ 's can be placed.

That is, 3 possibilities for the first $r, 2$ possibilities for the second $r$, and then the third $r$ is fixed.

The total 210 counts the number of ways to place the three coloured $r$ 's in the seven positions. We only care about the position of the $r$ 's, not the colour.

Each possible positioning of the $r$ 's will be included 6 times in the total 210 , so the number of ways to place three $r$ 's in seven positions is $\frac{201}{6}=35$.

Thus, the number of paths from $C$ to $Y$ is $\frac{210}{6}=35$.

We can now compute $f(C)=10 \times 35=350$.

The sort of reasoning used to compute $f(C)$ can be used to count paths from $X$ to $Y$ through the remaining points.

$f(D)$ : We first count the number of paths from $X$ to $D$.

This is the same as counting the number of sequences of five $r$ 's and two $d$ 's.

Similar to the reasoning used already, this is the same as counting the number of ways to place two d's in 7 positions.

The number of such sequences is $\frac{7 \times 6}{2}=21$.

The number of paths from $D$ to $Y$ is equal to 5 . This is equal to the number of sequences of one $r$ and four $d$ 's (there are five possible positions for the $r$ and the $d$ 's are fixed after placing the $r$ ).

This means $f(D)=21 \times 5=105$.

$\underline{f(E)}$ : The number of paths from $X$ to $E$ is 4 . The number of paths from $E$ to $Y$ is equal to the number of ways to place three $d$ 's in 8 positions, which is $\frac{8 \times 7 \times 6}{6}=56$. Therefore, $f(E)=4 \times 56=224$.

$\underline{f(F)}$ : The number of paths from $X$ to $F$ is equal to the number of paths from $F$ to $Y$ by symmetry.

The number of paths from $X$ to $F$ is equal to the number of ways to place three $r$ 's in six positions (or three d's in 6 positions), which is $\frac{6 \times 5 \times 4}{6}=20$.

This means $f(F)=20 \times 20=400$.

$f(G)$ : Again by symmetry, the number of paths from $X$ to $G$ is equal to the number of paths from $G$ to $Y$.

The number of paths from $X$ to $G$ is equal to $\frac{6 \times 5}{2}=15$, so $f(G)=15 \times 15=225$.

We have $f(A)=462, f(B)=1, f(C)=350, f(D)=105, f(E)=224, f(F)=400$, and $f(G)=225$. Therefore, the points listed in order from most paths to fewest paths is $A F C G E D B$.

ANSWER: $A F C G E D B$