Verified Solution • 2025–26 Session moderate

QJEEAdv08Ma60c8: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Ma60c8 Solution: Consider all possible permutations of the letters of the word…

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.

Column I Column II
(A) The number of permutations containing the word ENDEA is (P) 5!
(B) The number of permutations in which the letter E occurs in the first and the last position is (Q) 2 ×\times 5!
(C) The number of permutations in which none of the letters D, L, N occurs in the last five positions is (R) 7 ×\times 5!
(D) The number of permutations in which the letters A, E, O occur only in odd positions is (S) 21 ×\times 5!

Multiple Choice Options

  • A (A) - p ; (B) - s; (C) - q ; (D) - q
  • B (A) - q ; (B) - q ; (C) - s ; (D) - p
  • C (A) - p ; (B) - s; (C) - p ; (D) - r
  • D (A) - p ; (B) - r ; (C) - q ; (D) - p

Step-by-Step Verified Solution

(A) Considering ENDEA as one group, remaining letters are N, O, E, L.

So, no. of permutations = 5!

(A) - (i)

(B) E occurs in 1st and last positions. The remaining letters are N, N, D, A, O, E, L.

No. of permutation

=7!2!=7×62×5!=21!×5!= {{7!} \over {2!}} = {{7 \times 6} \over 2} \times 5! = 21! \times 5!

(B) - (iv)

(C) D, L, N should not occur in last five positions

⇒\Rightarrow D, L, N should occur in 1st four positions, but we have D, L, N, N.

So, ways of arranging D, L, N, N in 1st four positions

=4!2!=12= {{4!} \over {2!}} = 12

Ways of arranging remaining E, E, A, O, E in last five positions =5!3!=20= {{5!} \over {3!}} = 20 ,

Total =12×20=240=2×5!=12\times20=240=2\times5!

(C) - (iv)

(D) A, E, O occur in odd positions.

No. of odd positions = 5 and letters are E, E, E, A, O. i.e. 5

Ways of arranging those 5 letters in 5 odd positions =5!3!=20= {{5!} \over {3!}} = 20

Remaining 4 letters D, L, N, N can be arranged in remaining 4 positions in =4!2!=12= {{4!} \over {2!}} = 12 ways

Total no. of permutations =20×12=240=2×5!=20\times12=240=2\times5!

(D) - (iii)

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