Consider all possible permutations of...

MathematicsLinear PermutationsJEE Advanced 2008Moderate

View in:English Hindi

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 $4 \times 4$ 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 \times 5!

(3)

7 \times 5!

(4)

21 \times 5!

Answer:P → 1, Q → 4, R → 2, S → 2

Visualized Solution (English)

Word Analysis

Word: ENDEANOEL
Total letters: $9$
Letter counts: $E : 3, N : 2, D : 1, A : 1, O : 1, L : 1$

Part (A): Word ENDEA as a Block

Treat ENDEA as a single block.
Remaining letters: $N, O, E, L$
Total units to arrange: $1 (block) + 4 (letters) = 5$
Number of ways = $5!$
Matching: $(A) \to (p)$

Part (B): Fixed E at Ends

Fixed positions: $E_______E$
Remaining letters: $N, D, E, A, N, O, L$ (Total $7$ )
Repetitions: $N$ occurs $2$ times.
Arrangements = $\frac{7 !}{2 !} = \frac{7 \times 6 \times 5 !}{2} = 21 \times 5!$
Matching: $(B) \to (s)$

Part (C): Restrictions on Last 5 Slots

Constraint: $D, L, N, N$ cannot be in positions $5, 6, 7, 8, 9$ .
Therefore, $D, L, N, N$ must occupy positions $1, 2, 3, 4$ .
Arrangement of first 4: $\frac{4 !}{2 !} = 12$
Remaining letters ( $E, E, E, A, O$ ) in last 5: $\frac{5 !}{3 !} = 20$
Total = $12 \times 20 = 240 = 2 \times 5!$
Matching: $(C) \to (q)$

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

Odd positions: $1, 3, 5, 7, 9$ ( $5$ slots)
Letters for odd slots: $A, E, E, E, O$ (Total $5$ )
Arrangement (Odd): $\frac{5 !}{3 !} = 20$
Letters for even slots ( $2, 4, 6, 8$ ): $N, N, D, L$
Arrangement (Even): $\frac{4 !}{2 !} = 12$
Total = $20 \times 12 = 240 = 2 \times 5!$
Matching: $(D) \to (q)$

Final Conclusion

Final Matchings:
$(A) \to (p)$
$(B) \to (s)$
$(C) \to (q)$
$(D) \to (q)$
Key Takeaway: Handle restricted positions or blocks first before arranging the rest.

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List-I

List-II

Visualized Solution (English)

Word Analysis

Part (A): Word ENDEA as a Block

Part (B): Fixed E at Ends

Part (C): Restrictions on Last 5 Slots

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

Final Conclusion

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