Question

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.
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 # occurs in the first and the last positions is (q) $2\times 5!$
(C)The number of permutations in which none of the lettersD,L,N,occurs in the last five jpositions is (r) $7\times 5!$
(D) The number of pernutations in which the leters A, E,O occur only in odd positions is (s)$21\times 5!$

• A. The number of permutations containing the word ENDEA is (p)5!
• B. The number of jpermutations in which the letter # occurs in the first and the last positions is (q) $2\times 5!$
• C. The number of permutations in which none of the lettersD,L,N,occurs in the last five jpositions is (r) $7\times 5!$
• D. The number of pernutations in which the letersA, E,O occur only in odd positions is (s) $21\times 5!$

Answer 1

Answer: (A), (B), (C), (D)

The number of pernutations in which the letersA, E,O occur only in odd positions is (s) $21\times 5!$

For the permutations containing the word ENDEA we consider 'ENDEA' as single letter. Then we have total ENDEA,N,O,E,L i.e. 5 letters which can be arranged in 5! ways.
$\therefore \left(A\right)\to \left(p\right)$
If E occupies the first and last position, the middle 7 positions can be filled by N,E,A,N,O,L.
these can be arranged in $\frac{7!}{2!}=21\times 5!ways.$
$\therefore \left(B\right)\to \left(s\right)$
If none of the letters D,L,N occur in the last five positions then we should arrange D,D,L,N,at first four positions and rest five i.e.E, E, E, A,O at last five positions, This can be done in
$\frac{4!}{2!}\times \frac{5!}{3!}ways.\left(C\right)\to \left(q\right)$
As per question A,E,E, E, O can be aranged at lst, erd, 5th, 7th and 9th positions and rest N,D,N,L at rest 4 positions. This can be done in
$\frac{5!}{3!}\times \frac{4!}{2!}ways=2\times 5!ways\left(D\right)\to \left(q\right)$