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: (Figure: 3 rows of boxes, arranged in a T-shape, with a total of 8 boxes)

Step 1: Analyze the problem and constraints

We have 5 distinct letters (A, B, C, D, E) to place into 8 distinct boxes. The key constraints are that each box can hold at most one letter, and no row can remain empty. The figure shows 3 rows: Row 1 has 3 boxes, Row 2 has 2 boxes, and Row 3 has 3 boxes.

Step 2: Select 5 boxes out of 8

Since we have 5 letters and each box can hold at most one letter, we must choose exactly 5 boxes out of the 8 available boxes to place the letters. The number of ways to do this is given by the combination formula $\binom{n}{k}$.

Step 3: Arrange the 5 letters in the selected 5 boxes

Once 5 boxes are selected, the 5 distinct letters can be arranged in these 5 boxes in $5!$ ways. This is a permutation of 5 distinct items.

Step 4: Calculate total ways without the 'no row empty' constraint

The total number of ways to place 5 distinct letters into 8 distinct boxes such that each box has at most one letter is the product of selecting 5 boxes and then arranging the 5 letters in them.

Step 5: Identify and subtract invalid arrangements (rows empty)

We must subtract cases where a row remains empty. The rows have 3, 2, and 3 boxes. If Row 1 (3 boxes) or Row 3 (3 boxes) is empty, we would need to place 5 letters into the remaining 5 boxes (2+3=5 or 3+2=5). This is possible. If Row 2 (2 boxes) is empty, we would need to place 5 letters into the remaining 6 boxes (3+3=6). This is also possible. However, if we place 5 letters into 8 boxes, it is impossible for two rows to be empty, as that would leave at most 3 boxes (if Row 1 and Row 3 are empty, only Row 2 with 2 boxes remains, which cannot hold 5 letters). If Row 1 and Row 2 are empty, only Row 3 with 3 boxes remains. If Row 2 and Row 3 are empty, only Row 1 with 3 boxes remains. In all these cases, we cannot place 5 letters. So, the only invalid case is when Row 2 (the middle row with 2 boxes) is empty. If Row 2 is empty, all 5 letters must be placed in the remaining $3+3=6$ boxes. The number of ways to choose 5 boxes from these 6 boxes is $\binom{6}{5} = 6$. The number of ways to arrange 5 letters in these 5 boxes is $5! = 120$. So, the number of ways where Row 2 is empty is $6 \times 120 = 720$. Wait, let's re-evaluate. The problem states 'no row remains empty'. This means we must ensure each of the 3 rows has at least one letter. The total number of boxes is 8. The rows are $R_1=3, R_2=2, R_3=3$. We are placing 5 letters. It is impossible for $R_1$ to be empty, because if $R_1$ is empty, we have $2+3=5$ boxes left in $R_2$ and $R_3$. We must place 5 letters in these 5 boxes. This means $R_2$ and $R_3$ are full. So, if $R_1$ is empty, then $R_2$ and $R_3$ are full. This is a valid scenario where $R_1$ is empty. Similarly, if $R_3$ is empty, then $R_1$ and $R_2$ are full. This is also a valid scenario where $R_3$ is empty. If $R_2$ is empty, then $R_1$ and $R_3$ must contain the 5 letters. This means we choose 5 boxes from $3+3=6$ boxes. This is $\binom{6}{5} = 6$ ways to choose the boxes, and $5! = 120$ ways to arrange the letters. So, $6 \times 120 = 720$ ways where $R_2$ is empty. Let's use the principle of inclusion-exclusion. Total ways = 6720. Cases where $R_1$ is empty: We choose 5 boxes from $R_2 \cup R_3$ (5 boxes). $\binom{5}{5} \times 5! = 1 \times 120 = 120$. Cases where $R_2$ is empty: We choose 5 boxes from $R_1 \cup R_3$ (6 boxes). $\binom{6}{5} \times 5! = 6 \times 120 = 720$. Cases where $R_3$ is empty: We choose 5 boxes from $R_1 \cup R_2$ (5 boxes). $\binom{5}{5} \times 5! = 1 \times 120 = 120$. Cases where two rows are empty: Not possible to place 5 letters. For example, if $R_1$ and $R_2$ are empty, only $R_3$ (3 boxes) remains, which cannot hold 5 letters. So, the number of invalid ways is $120 + 720 + 120 = 960$. Therefore, the number of ways is $6720 - 960 = 5760$.

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