The given word is ENDEANOEL. We need to find the number of permutations based on different conditions.

First, let's analyze the letters in the word ENDEANOEL: Total number of letters = 9. The letters are: E (3 times), N (2 times), D (1 time), A (1 time), O (1 time), L (1 time).

Let's evaluate each statement in Column I.

(A) The number of permutations containing the word ENDEA is

1. We can treat the word "ENDEA" as a single block or a single unit. Let's denote this block by (X).
2. The letters used to form the block "ENDEA" are E, N, D, E, A. The original letters of ENDEANOEL are {E, E, E, N, N, D, A, O, L}.
3. After forming the block, the remaining letters are {E, N, O, L}.
4. We now need to arrange the 5 items: {(ENDEA), E, N, O, L}. These 5 items are all distinct.
5. The number of permutations of these 5 distinct items is $5!$.
6. $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
7. This matches with (P) in Column II, which is $5!$. Therefore, (A) -> (P).

(B) The number of permutations in which the letter E occurs in the first and the last position is

1. The word has 9 positions. We fix the letter E at the first and the last positions. E _ _ _ _ _ _ _ E
2. We have used two of the three E's. The remaining 7 letters are {E, N, N, D, A, O, L}.
3. These 7 letters need to be arranged in the 7 middle positions.
4. In this set of 7 letters, 'N' is repeated 2 times.
5. The number of permutations is given by $\frac{7!}{2!} = \frac{5040}{2} = 2520$.
6. We need to express this result in terms of $5!$ ($5! = 120$). $2520 = 21 \times 120 = 21 \times 5!$.
7. This matches with (S) in Column II, which is $21 \times 5!$. Therefore, (B) -> (S).

(C) The number of permutations in which none of the letters D, L, N occurs in the last five positions is

1. There are 9 positions in total. We can conceptually divide them into two sets: the first four positions and the last five positions.
2. The letters specified are D, L, and N. In the word ENDEANOEL, we have D, L, N, N. These are 4 letters.
3. The condition states that these 4 letters (D, L, N, N) cannot be in the last five positions. This implies they must occupy the first four positions.
4. The remaining 5 letters are {E, E, E, A, O}. These letters must occupy the last five positions.
5. First, let's find the number of ways to arrange {D, L, N, N} in the first four positions. Since 'N' is repeated twice, the number of ways is $\frac{4!}{2!} = \frac{24}{2} = 12$.
6. Next, let's find the number of ways to arrange {E, E, E, A, O} in the last five positions. Since 'E' is repeated thrice, the number of ways is $\frac{5!}{3!} = \frac{120}{6} = 20$.
7. The total number of such permutations is the product of these two values: $12 \times 20 = 240$.
8. Expressing 240 in terms of $5!$: $240 = 2 \times 120 = 2 \times 5!$.
9. This matches with (Q) in Column II, which is $2 \times 5!$. Therefore, (C) -> (Q).

(D) The number of permutations in which the letters A, E, O occur only in odd positions is

1. In a 9-letter arrangement, there are 5 odd positions (1, 3, 5, 7, 9) and 4 even positions (2, 4, 6, 8).
2. The letters A, E, O refer to all occurrences of these letters in the word. The set of vowels is {A, E, E, E, O}, which are 5 letters in total.
3. The set of consonants is {N, N, D, L}, which are 4 letters in total.
4. The condition requires that the 5 vowels {A, E, E, E, O} must be placed in the 5 odd positions.
5. Consequently, the 4 consonants {N, N, D, L} must be placed in the 4 even positions.
6. The number of ways to arrange the vowels in the 5 odd positions is $\frac{5!}{3!} = \frac{120}{6} = 20$ (since E is repeated 3 times).
7. The number of ways to arrange the consonants in the 4 even positions is $\frac{4!}{2!} = \frac{24}{2} = 12$ (since N is repeated 2 times).
8. The total number of such permutations is the product: $20 \times 12 = 240$.
9. Expressing 240 in terms of $5!$: $240 = 2 \times 120 = 2 \times 5!$.
10. This matches with (Q) in Column II, which is $2 \times 5!$. Therefore, (D) -> (Q).

Summary of Matches: (A) -> (P) (B) -> (S) (C) -> (Q) (D) -> (Q)

This corresponds to the option: (A) - p ; (B) - s; (C) - q ; (D) - q.