JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 17)
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! |
Explanation
(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
$$ = {{7!} \over {2!}} = {{7 \times 6} \over 2} \times 5! = 21! \times 5!$$
(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
$$ = {{4!} \over {2!}} = 12$$
Ways of arranging remaining E, E, A, O, E in last five positions $$ = {{5!} \over {3!}} = 20$$,
Total $$=12\times20=240=2\times5!$$
(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 $$ = {{5!} \over {3!}} = 20$$
Remaining 4 letters D, L, N, N can be arranged in remaining 4 positions in $$ = {{4!} \over {2!}} = 12$$ ways
Total no. of permutations $$=20\times12=240=2\times5!$$
(D) - (iii)
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