Match Permutation Counts of Word ENDEANOEL with Given Conditions

To determine the correct matches between Column I and Column II, let us first analyze the letters in the word ENDEANOEL and their respective frequencies:

• Total letters: 9
• Letter frequencies:
  • $\text{E} : 3$
  • $\text{N} : 2$
  • $\text{D} : 1$
  • $\text{A} : 1$
  • $\text{O} : 1$
  • $\text{L} : 1$

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Analysis of Statement (A):

The number of permutations containing the word $\text{ENDEA}$

We can treat the word $\text{ENDEA}$ as a single block/unit $X = [\text{ENDEA}]$.

• Letters used in $X$: $\text{E (2), N (1), D (1), A (1)}$.
• Remaining letters: $\text{E (1), N (1), O (1), L (1)}$.

Now, we need to arrange the block $X$ along with the remaining 4 letters: $\{X, \text{E}, \text{N}, \text{O}, \text{L}\}$.
Since all 5 elements are distinct, the number of arrangements is:
$\text{Number of permutations} = 5!$

Thus, (A) matches with (p).

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Analysis of Statement (B):

The number of permutations in which the letter $\text{E}$ occurs in the first and the last positions

• Fix $\text{E}$ in the $1^{\text{st}}$ position and $\text{E}$ in the $9^{\text{th}}$ position.
• This leaves 7 middle positions (from position $2$ to $8$) to be filled by the remaining 7 letters: $\{\text{E (1), N (2), D (1), A (1), O (1), L (1)}\}$.

The number of ways to arrange these 7 letters (where $\text{N}$ repeats 2 times) is:
$\text{Number of permutations} = \frac{7!}{2!} = \frac{7 \times 6 \times 5!}{2} = 21 \times 5!$

Thus, (B) matches with (s).

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Analysis of Statement (C):

The number of permutations in which none of the letters $\text{D, L, N}$ occurs in the last five positions

The 9 positions in the permutation are $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. The last five positions are $\{5, 6, 7, 8, 9\}$.

• The letters $\text{D, L, N}$ consist of 4 letters in total: $\{\text{D (1), L (1), N (2)}\}$.
• Since none of these 4 letters can occupy the last 5 positions, they must occupy the first 4 positions $\{1, 2, 3, 4\}$.

1. Number of ways to arrange $\{\text{D}, \text{L}, \text{N}, \text{N}\}$ in the first 4 positions:
   $\frac{4!}{2!} = 12$

2. The remaining 5 letters $\{\text{E (3), A (1), O (1)}\}$ must occupy the last 5 positions. The number of ways to arrange them is:
   $\frac{5!}{3!} = 20$

Total permutations:
$\text{Number of permutations} = 12 \times 20 = 240 = 2 \times 5!$

Thus, (C) matches with (q).

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Analysis of Statement (D):

The number of permutations in which the letters $\text{A, E, O}$ occur only in odd positions

• Total positions = 9.
• Odd positions = $\{1, 3, 5, 7, 9\}$ (5 positions in total).
• Even positions = $\{2, 4, 6, 8\}$ (4 positions in total).

The letters $\{\text{A, E, O}\}$ consist of 5 letters in total: $\{\text{A (1), E (3), O (1)}\}$.
Since these 5 letters can occur only in odd positions, they must occupy all 5 odd positions.

1. Number of ways to arrange $\{\text{A}, \text{E}, \text{E}, \text{E}, \text{O}\}$ in the 5 odd positions:
   $\frac{5!}{3!} = 20$

2. The remaining 4 letters $\{\text{N (2), D (1), L (1)}\}$ must occupy the 4 even positions. Number of ways to arrange them:
   $\frac{4!}{2!} = 12$

Total permutations:
$\text{Number of permutations} = 20 \times 12 = 240 = 2 \times 5!$

Thus, (D) matches with (q).

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Conclusion:

• $(\text{A}) \rightarrow (\text{p})$
• $(\text{B}) \rightarrow (\text{s})$
• $(\text{C}) \rightarrow (\text{q})$
• $(\text{D}) \rightarrow (\text{q})$