AMC 10B 2024 (Problem 20)
Three different pairs of shoes are placed in a row so that no left shoe is next to a right shoe from a different pair. In how many ways can these six shoes be lined up?
Let the three pairs of shoes be , , and .
Without loss of generality, assume the first shoe is . At the end, multiply the number of arrangements by .
The second shoe can be , , or .
The cases in which the second shoe is or are symmetric.
The cases in which the second shoe is or are symmetric.
Case 1: The first two shoes are . There are symmetric choices for the second shoe.
Case 2: The first two shoes are .
Case 2: The first two shoes are .
Case 1: The first two shoes are .
Case 1-1:
Case 1-2:
Case 1-1:
Case 1-2:
Case 1-1:
The last two shoes can be or .
Therefore, this case gives arrangements.
The last two shoes can be or .
Therefore, this case gives arrangements.
Case 1-2:
This gives valid arrangement.
The following arrangements are invalid:
This gives valid arrangement.
The following arrangements are invalid:
Case 2: The first two shoes are .
There are symmetric choices for the third shoe. Assume it is .
Case 2-1:
Case 2-2:
There are symmetric choices for the third shoe. Assume it is .
Case 2-1:
Case 2-2:
Case 2-1:
This gives arrangement.
This gives arrangement.
Case 2-2:
This gives arrangement.
This gives arrangement.
Combining all the cases and accounting for the choices of the first shoe, we obtain
(A) 60
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