Get the complete, step-by-step math solution for: "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 l...". Powered by SolveForX AI math tutor.
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 (kn).
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 (56)=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 6×120=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 R1=3,R2=2,R3=3. We are placing 5 letters. It is impossible for R1 to be empty, because if R1 is empty, we have 2+3=5 boxes left in R2 and R3. We must place 5 letters in these 5 boxes. This means R2 and R3 are full. So, if R1 is empty, then R2 and R3 are full. This is a valid scenario where R1 is empty. Similarly, if R3 is empty, then R1 and R2 are full. This is also a valid scenario where R3 is empty. If R2 is empty, then R1 and R3 must contain the 5 letters. This means we choose 5 boxes from 3+3=6 boxes. This is (56)=6 ways to choose the boxes, and 5!=120 ways to arrange the letters. So, 6×120=720 ways where R2 is empty. Let's use the principle of inclusion-exclusion. Total ways = 6720. Cases where R1 is empty: We choose 5 boxes from R2∪R3 (5 boxes). (55)×5!=1×120=120. Cases where R2 is empty: We choose 5 boxes from R1∪R3 (6 boxes). (56)×5!=6×120=720. Cases where R3 is empty: We choose 5 boxes from R1∪R2 (5 boxes). (55)×5!=1×120=120. Cases where two rows are empty: Not possible to place 5 letters. For example, if R1 and R2 are empty, only R3 (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.