The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:
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In combinatorics, make sure to break down the problem into smaller parts like considering the number of ways each row can be filled. Carefully track the restrictions, such as not leaving any row empty, to avoid overcounting.
Let the 8 boxes be arranged in three rows as shown: Let \( R_1, R_2, R_3 \) represent the three rows. \( R_1 \to \) (1st row), \( R_2 \to \) (2nd row), \( R_3 \to \) (3rd row). Total number of ways: \[ \text{Total} = \left[ (\text{All in } R_1 \text{ and } R_3) + (\text{All in } R_2 \text{ and } R_3) + (\text{All in } R_1 \text{ and } R_2) \right] \] \[ = 8C5 \times 5! - \left[ \text{(ways to place in 1st and 2nd row)} + \text{(ways to place in 3rd row)} \right] \] \[ = \left| (56-1) \times 6 \right| = 120 \times 48 = 5760 \] Hence, the total number of ways to arrange the letters is \( 5760 \).
