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:

### Related Formula textNumber of arrangements of r text items in n text boxes = binomnr cdot r!$$\text{Number of arrangements of } r \text{ items in } n \text{ boxes} = \binom{n}{r} \cdot r!$$ ### Core Logic This is a permutations problem with row constraints. We compute the total arrangements of placing 5 distinct letters into 8 boxes and then subtract the invalid cases where one or more rows are left completely empty. ### Step 1: Compute total unrestricted arrangements The grid has a total of 8 boxes. We have 5 distinct letters (A, B, C, D, E): textTotal unrestricted arrangements = binom85 cdot 5! = 56 cdot 120 = 6720$$\text{Total unrestricted arrangements} = \binom{8}{5} \cdot 5! = 56 \cdot 120 = 6720$$ ### Step 2: Identify and subtract the invalid empty-row cases Let the rows be R_1$R_1$, R_2$R_2$, and R_3$R_3$, with box counts 3, 2, and 3 respectively. Since we must distribute 5 letters, it is impossible for 2 rows to be empty simultaneously (as the remaining single row would have at most 3 boxes, which cannot fit 5 letters). Thus, we only subtract cases where exactly one row is empty: - Case 1: Row R_1$R_1$ (3 boxes) is empty. The 5 letters must go to the remaining 5 boxes of R_2$R_2$ and R_3$R_3$: textWays = binom55 cdot 5! = 120$$\text{Ways} = \binom{5}{5} \cdot 5! = 120$$ - Case 2: Row R_3$R_3$ (3 boxes) is empty. Same as Case 1, the 5 letters must go to the remaining 5 boxes of R_1$R_1$ and R_2$R_2$: textWays = binom55 cdot 5! = 120$$\text{Ways} = \binom{5}{5} \cdot 5! = 120$$ - Case 3: Row R_2$R_2$ (2 boxes) is empty. The 5 letters must go to the remaining 6 boxes of R_1$R_1$ and R_3$R_3$: textWays = binom65 cdot 5! = 6 cdot 120 = 720$$\text{Ways} = \binom{6}{5} \cdot 5! = 6 \cdot 120 = 720$$ ### Step 3: Calculate the final valid arrangements Subtracting all empty-row cases from the total arrangements: textValid arrangements = 6720 - (120 + 120 + 720) = 6720 - 960 = 5760$$\text{Valid arrangements} = 6720 - (120 + 120 + 720) = 6720 - 960 = 5760$$ ### Pattern Recognition Inclusion-Exclusion Principle: For distribution problems with simple boundary exclusions, subtracting the complement set (invalid configurations) is mathematically much cleaner than calculating all possible partitions of row assignments. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations