Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements / Expr

Q: Wait, why do we divide by $n!$ when there are repeated letters?

Great question! Think of it this way: if all letters were distinct, every permutation would be unique. But when you have identical letters (like the three E's in ENDEANOEL), swapping two identical E's doesn't create a new, distinct word. Dividing by the factorial of the number of repetitions effectively 'divides out' these redundant, identical-looking arrangements, leaving you with only the truly distinct permutations.

Q: Why do we treat 'ENDEA' as a single block in Part A?

The 'string method' or 'block method' is a powerful tool in combinatorics. By tying E, N, D, E, A together, we ensure they act as one inseparable unit. This forces them to stay in that exact order relative to each other. Once you treat them as one block, you just arrange that block along with the other remaining letters, simplifying the problem significantly.

Q: Why do we multiply the results for the front and back sections in Part C?

This is the 'Fundamental Counting Principle' in action. Since the arrangement of the first four slots and the arrangement of the last five slots are independent events—meaning the choice of letters in the front doesn't restrict the choices in the back, given the constraints—we multiply the number of ways to complete each task to find the total number of ways to complete the entire process.

Question

Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements / Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 	imes 4$ matrix given in the ORS.

List-I:

P. The number of permutations containing the word ENDEA is

Q. The number of permutations in which the letter E occurs in the first and the last positions is

R. The number of permutations in which none of the letters D, L, N occurs in the last five positions is

S. The number of permutations in which the letters A, E, O occur only in odd positions is

List-II:

1. $5!$

2. $2 	imes 5!$

3. $7 	imes 5!$

4. $21 	imes 5!$

Sigma Percentile · Accepted Answer

### Analyzing the Word Structure

To solve these problems, we first perform an autopsy on the word **ENDEANOEL**. There are a total of $9$ letters.

The frequency distribution of the letters is as follows:
*   $E$: $3$ times
*   $N$: $2$ times
*   $D, A, O, L$: $1$ time each

This frequency distribution serves as our map for all subsequent calculations.

### Part (A): Permutations containing the block ENDEA

We employ the **block method** here. By tying $E, N, D, E, A$ into one unbreakable unit, we are left with this block plus the remaining letters: $N, O, E, L$.

This gives us $5$ units to arrange. Since the block acts as a single unit, the number of ways is simply:
$$5! = 120$$

This result matches option **(p)**.

### Part (B): Permutations with E at the first and last positions

We fix $E$ at the first and last positions. This leaves $7$ slots in the middle to be filled by the remaining letters: $N, D, E, A, N, O, L$.

Among these $7$ letters, $N$ repeats twice. The number of arrangements is:
$$\frac{7!}{2!} = \frac{5040}{2} = 2520$$

Expanding this, we get:
$$\frac{7 	imes 6 	imes 5!}{2} = 21 	imes 120 = 2520$$

This matches option **(s)**.

### Part (C): Constraints on D, L, N, N

We are told that $D, L, N, N$ cannot be in the last $5$ positions. This forces them into the first $4$ positions.

The number of ways to arrange these $4$ letters ($D, L, N, N$) in the first $4$ slots is:
$$\frac{4!}{2!} = 12$$

The remaining $5$ slots must be filled by the remaining letters: $E, E, E, A, O$. The number of ways to arrange these is:
$$\frac{5!}{3!} = 20$$

Multiplying these independent events, we get $12 	imes 20 = 240$, which is $2 	imes 5!$. This matches option **(q)**.

### Part (D): Permutations with Vowels in Odd Positions

There are $5$ odd positions ($1, 3, 5, 7, 9$) and $5$ vowels ($A, E, E, E, O$). The number of ways to arrange these is:
$$\frac{5!}{3!} = 20$$

The remaining $4$ even positions ($2, 4, 6, 8$) are filled by the remaining letters ($N, N, D, L$). The number of ways to arrange these is:
$$\frac{4!}{2!} = 12$$

Again, the total number of arrangements is $20 	imes 12 = 240$, or $2 	imes 5!$. This also matches option **(q)**.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Word Analysis

Part (A): Word ENDEA as a Block

Part (A): Calculation

Part (B): Fixed E at Ends

Part (B): Calculation

Part (C): Restrictions on Last 5 Slots

Part (C): Placing in First 4 Slots

Part (C): Calculation for First 4

Part (C): Calculation for Last 5

Part (C): Total Permutations

Part (D): A, E, O at Odd Positions

Part (D): Arranging in Odd Positions

Part (D): Arranging in Even Positions

Part (D): Total Permutations

Final Matchings

The Sigma Insight: Linear Permutations

Analyzing the Word Structure

Part (A)

Part (B)

Part (C)

Part (D)

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