Consider folding the net into the cube and positioning the cube with the face labelled $F$ down (on the bottom), and so the face labelled $A$ is on top and the four remaining vertical faces are as shown.

Beginning at $A$, the ant has 4 choices for which face to visit next.

That is, the ant can walk from $A$ to any of the vertical faces, $B, C$,

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-038.jpg?height=290&width=328&top_left_y=1774&top_left_x=1560)
$D$, or $E$.

From each of these vertical faces, the ant can walk to $F$ (the bottom face), or the ant can walk to an adjacent vertical face.

We call these two possibilities Case 1 and Case 2, and for each case we consider the number of possible orders in which the ant can visit the faces.

Case 1: The ant's 2nd move is to $F$, the bottom face

We begin by recognizing that there is only 1 choice for the ant's 2 nd move in this case.

From $F$, the ant's 3rd move must be back to a vertical face.

The ant cannot return to the vertical face that it has already visited.

Also, the ant cannot move to the vertical face that is opposite the vertical face it has already
visited. Why?

Consider for example that the ant visits, in order, $A, B, F$.

If the ant walks to $E$ (the face opposite $B$ ) on its 3rd move, then its 4 th move must be to $C$ or to $D$ (since it has visited the other four faces).

However, once at $C$ or $D$, the ant is "trapped" since it has already visited all four adjacent faces, and thus cannot get to the sixth face.

Thus from $F$, the ant's 3rd move must be to a vertical face that is not the face already visited, and is not the face opposite the face already visited, and so there are 2 choices for the next move.

From this vertical face, the ant must walk to another vertical face (since it has already visited $A$ and $F$ ).

One such adjacent face has already been visited and the other has not, and so the ant's 1 choice is for its 4th move to be to the adjacent vertical face it has not visited.

For example, if the order is $A, B, F, C$, then the ant must move to $E$ next.

Finally, the ant's final move must be to the adjacent vertical face that it has not visited.

Summarizing Case 1, there are 4 choices for the first move from $A$ to one of the vertical faces, 1 choice for the 2nd move (to $F$ ), 2 choices for the 3rd move, and 1 choice for each of the last two moves, and so there are $4 \times 1 \times 2 \times 1 \times 1=8$ possible orders.

Case 2: The ant's 2nd move is to an adjacent vertical face

There are two vertical faces adjacent to each vertical face, and so the ant has 2 choices for its 2nd move.

For its 3rd move, the ant can walk to the bottom face $F$, or the ant can walk to the adjacent face that has not been visited.

We call the first of these Case 2a and the second Case 2b.

Case 2a: The ant's 3rd move is to $F$

We begin by recognizing that there is only 1 choice for the ant's 3rd move in this case.

From $F$, the ant's 4 th move must be back to a vertical face.

The ant cannot return to either of the two vertical faces that it has already visited, and so there are 2 choices for the ant's 4th move.

For example, if the order is $A, B, C, F$, the ant's 4th move can be to $D$ or to $E$. The ant's final move is to the final vertical face and thus there is only 1 choice.

Summarizing Case 2a, there are 4 choices for the first move from $A$ to one of the vertical faces, 2 choices for the 2nd move (to an adjacent vertical face), 1 choice for the 3rd move (to $F$ ), 2 choices for the 4th move, and 1 choice for the last move, and so there are $4 \times 2 \times 1 \times 2 \times 1=16$ possible orders.

Case 2b: The ant's 3rd move is to the adjacent face that has not been visited

We begin by recognizing that there is only 1 choice for the ant's 3rd move in this case.

At this point the ant has visited three vertical faces.

The ant's 4 th move can be to $F$ or to the final vertical face.

That is, the ant has 2 choices for its 4th move.

If the ant's 4th move is to $F$, then its last move is to the remaining vertical face.

If the ant's 4th move is to the final vertical face, then its last move is to $F$.

That is, once the ant chooses its 4th move, it has only 1 choice for its final move.

Summarizing Case $2 \mathrm{~b}$, there are 4 choices for the first move from $A$ to one of the
vertical faces, 2 choices for the 2nd move (to an adjacent vertical face), 1 choice for the 3rd move (to the adjacent vertical face), 2 choices for the 4th move (to $F$ or to the final vertical face), and 1 choice for the last move, and so there are $4 \times 2 \times 1 \times 2 \times 1=16$ possible orders.

Thus, if the ant starts at $A$ and visits each face exactly once, there are $8+16+16=40$ possible orders.

ANSWER: (E)

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The net shown is folded to form a cube. An ant walks from face to face on the cube, visiting each face exactly once. For example, $A B C F E D$ and $A B C E F D$ are two possible orders of faces the ant visits. If the ant starts at $A$, how many possible orders are there?
(A) 24
(B) 48
(C) 32
(D) 30
(E) 40

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-009.jpg?height=328&width=249&top_left_y=194&top_left_x=1407)