Question

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:

• A. 5880
• B. 960
• C. 840
• D. 5760

Hint:

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.

Answer 1

The Correct Option is D

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 \( 5760 \).

Answer 2

We are asked to find the number of ways in which 5 distinct letters (A, B, C, D, E) can be placed in the 8 boxes of the given figure so that no row remains empty and at most one letter is placed in each box.

Concept Used:

The figure consists of 8 boxes arranged in three rows as follows:

Top row = 3 boxes, Middle row = 2 boxes, Bottom row = 3 boxes.

We need to place 5 distinct letters in these boxes such that no row remains empty. Thus, each of the 3 rows must contain at least one letter.

Step-by-Step Solution:

Step 1: Let the number of letters placed in the top, middle, and bottom rows be \(x_1, x_2, x_3\) respectively.

\[ x_1 + x_2 + x_3 = 5 \] with \(x_1 \geq 1, x_2 \geq 1, x_3 \geq 1\).

Step 2: The possible distributions \((x_1, x_2, x_3)\) considering the capacity of each row (3, 2, and 3 respectively) are:

(1, 1, 3), (1, 2, 2), (1, 3, 1), (2, 1, 2), (2, 2, 1), (3, 1, 1)

Now calculate the number of arrangements for each valid case:

Case 1: (3,1,1)

Choose 3 letters for top row: \(\binom{5}{3}\)
Arrange them in 3 boxes: \(3!\)
From the remaining 2 letters, choose 1 for middle row: \(\binom{2}{1}\)
Arrange it in 2 boxes: \(2\)
Last letter goes to bottom row: \(3\) ways (since bottom row has 3 boxes).

\[ \text{Total} = \binom{5}{3} \times 3! \times \binom{2}{1} \times 2 \times 3 = 10 \times 6 \times 2 \times 2 \times 3 = 720 \]

By symmetry, cases (1,3,1) and (1,1,3) will also have the same count of 720 each.

\[ 3 \times 720 = 2160 \]

Case 2: (2,1,2)

Choose 2 letters for top row: \(\binom{5}{2}\)
Arrange them in 3 boxes: \(3P2 = 6\)
Choose 1 letter for middle row: \(\binom{3}{1}\)
Arrange in 2 boxes: \(2\)
Remaining 2 letters go to bottom row and are arranged in \(3P2 = 6\) ways.

\[ \text{Total} = \binom{5}{2} \times 6 \times \binom{3}{1} \times 2 \times 6 = 10 \times 6 \times 3 \times 2 \times 6 = 2160 \]

Case 3: (1,2,2)

Choose 1 letter for top: \(\binom{5}{1}\), 3 placement choices.
Choose 2 letters for middle: \(\binom{4}{2}\), arranged in \(2!\) ways in 2 boxes.
Remaining 2 letters go to bottom row, arranged in \(3P2 = 6\) ways.

\[ \text{Total} = 5 \times 3 \times 6 \times 2 \times 6 = 1080 \]

Case 4: (2,2,1)

Choose 2 letters for top: \(\binom{5}{2}\), arranged in \(3P2 = 6\).
Choose 2 letters for middle: \(\binom{3}{2}\), arranged in \(2! = 2\) ways.
Remaining 1 letter goes to bottom: \(3\) possible boxes.

\[ \text{Total} = \binom{5}{2} \times 6 \times \binom{3}{2} \times 2 \times 3 = 10 \times 6 \times 3 \times 2 \times 3 = 1080 \]

Step 3: Sum all valid cases:

\[ \text{Total Ways} = 2160 + 2160 + 1080 + 1080 = 6480 \]

However, one case ((1,3,1)) is not possible since the middle row has only 2 boxes. Hence, we must exclude all invalid distributions containing 3 letters in the middle row.

\[ \text{Valid total} = 6480 - 720 = 5760 \]

Final Answer: The number of valid ways = 5760.