2024 AMC 10A Problem 24

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled

A^+,

A^-,

B^+,

B^-,

C^+,

and

C^-

is rolled. Suppose the bee occupies the point

(a, b, c).

If the die shows

A^+,

then the bee moves to the point

(a + 1, b, c),

and if the die shows

A^-,

then the bee moves to the point

(a - 1, b, c).

Analogous moves are made with the other four outcomes. Suppose the bee starts at the point

(0, 0, 0)

and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?

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Solution

Every roll moves the bee one unit along

\pm x,\pm y,

\pm z,

so there are

6^4=1296

equally likely sequences. There are two types of valid paths. A path around one square face has

3

choices of coordinate plane,

4

choices for the signs of its two axes, and

2

choices for which axis is used first, giving

24.

Otherwise all three coordinate directions are used, with one repeated: choose that axis in

3

ways, choose its two nonadjacent positions in

3

ways, order the other two axes in

2

ways, and choose their three initial signs in

2^3=8

ways. (The second step on the repeated axis must have the opposite sign.) This gives

3\cdot3\cdot2\cdot8=144.

Hence there are

24+144=168

favorable sequences, and the probability is

\frac{168}{1296}=\frac{7}{54}.

Therefore, the answer is B.