The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure be

Q: Wait, why can't we have two rows empty simultaneously?

That is a brilliant observation! If two rows were empty, we would have at most 3 boxes left (from the remaining row). But we have 5 distinct letters to place. Since 5 is greater than 3, it is physically impossible to place all letters if two rows are empty. Therefore, the intersection of these 'bad' cases is zero.

Q: Why use permutations instead of combinations?

Great question. We use permutations because the letters A, B, C, D, and E are distinct. Placing 'A' in box 1 and 'B' in box 2 is a different arrangement than placing 'B' in box 1 and 'A' in box 2. Since the order of placement matters, permutations are the correct tool.

Q: Why is complementary counting better here?

Direct counting would require us to consider cases where 0 rows are empty, 1 row is empty, etc., which is complex. Complementary counting allows us to focus on the 'bad' scenarios—where a row is empty—and subtract them from the total universe of possibilities. It simplifies the logic significantly.

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 :

Options:

(A) 5880

(B) 960

(C) 840

(D) 5760

Sigma Percentile · Accepted Answer

### The Elegance of Exclusion: Solving the Grid Problem

Imagine you are standing before a grid of eight boxes, arranged in three distinct rows. You hold five unique letters—$A$, $B$, $C$, $D$, and $E$—in your hand.

Your mission is to place these letters into the boxes such that no row is left empty. At first glance, this might seem like a daunting puzzle of permutations and constraints.

As we dive into the mathematics, you will see that the most complex problems often yield to the most elegant strategies.

### The Total Universe

Before we worry about the constraints, let us consider the 'Total Universe' of possibilities. If there were no rules, how many ways could we place five distinct letters into eight available boxes?

Since each letter is unique and the order in which they occupy the boxes matters, we are dealing with a classic permutation problem. We are selecting and arranging five items out of eight, represented as ${}^{8}P_{5}$.

Calculating this is straightforward:

$$
{}^{8}P_{5} = \frac{8!}{(8-5)!} = 8 	imes 7 	imes 6 	imes 5 	imes 4 = 6720
$$

So, there are $6720$ ways to place the letters without any restrictions. This is our baseline.

### The Art of Subtraction

Now, we introduce the constraint: no row can remain empty. Trying to count the valid arrangements directly would be a nightmare of case-by-case analysis.

Instead, we use the power of **complementary counting**. We take our total universe of $6720$ and subtract the 'unwanted' cases—those where at least one row is empty.

### Analyzing the Unwanted Cases

Let us break down the unwanted scenarios. We have three rows, and we need to calculate the number of ways each row could be empty.

**Case 1: Row 1 is Empty**
Row 1 contains $3$ boxes. If we leave these empty, we are left with $8 - 3 = 5$ boxes.

We must place our $5$ letters into these $5$ boxes. The number of ways to do this is ${}^{5}P_{5}$, which is $5! = 120$.

**Case 2: Row 2 is Empty**
Row 2 also contains $3$ boxes. Similarly, if we leave these empty, we are left with $5$ boxes.

The number of ways to place our $5$ letters is again ${}^{5}P_{5} = 120$.

**Case 3: Row 3 is Empty**
Row 3 is different; it only contains $2$ boxes. If we leave these empty, we are left with $8 - 2 = 6$ boxes.

We must place our $5$ letters into these $6$ boxes. This is ${}^{6}P_{5}$, which is $6 	imes 5 	imes 4 	imes 3 	imes 2 = 720$.

### The Final Synthesis

We have identified our unwanted cases: $120$ (Row 1 empty), $120$ (Row 2 empty), and $720$ (Row 3 empty).

It is impossible for two rows to be empty simultaneously because we would not have enough boxes to hold our $5$ letters. Thus, the total number of unwanted arrangements is simply the sum of these three cases:

$$
	ext{Total Unwanted} = 120 + 120 + 720 = 960
$$

Finally, we subtract this from our total universe:

$$
	ext{Required Ways} = 6720 - 960 = 5760
$$

And there it is! By stepping back and looking at the problem through the lens of what we *don't* want, we have arrived at the solution with clarity and precision.

Remember, in combinatorics, the path of least resistance is often the one that leads to the most beautiful answer. The final result is **$5760$**.

Animated Solution for Mathematics - Permutations and Combinations: 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 :

Select Answer:

Visualized Solution

$Analyze the Grid Structure$

$Total Ways Without Constraints$

$Calculate^{8} P_{5}$

$Strategy: Complementary Counting$

$Case 1: Row 1 is Empty$

$Calculate Case 1$

$Case 2: Row 2 is Empty$

$Case 3: Row 3 is Empty$

$Calculate Case 3$

$Final Subtraction Setup$

$Conclusion and Final Answer$

The Sigma Insight: Linear Permutations

The Elegance of Exclusion

The Total Universe

The Art of Subtraction

Analyzing the Unwanted Cases

The Final Synthesis

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