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

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 positions is	q	$2 \times 5 !$
C	The number of permutations in which none of the letters $D, L, N$ occurs in the last five	r	$7 \times 5 !$ positions is
D	The number of permutations in which the letters $A , E , O$ occur only in odd positions is	s	$21 \times 5 !$

• A. $( A ) \rightarrow( P ) ;( B ) \rightarrow( S ) ;( C ) \rightarrow( Q ) ;( D ) \rightarrow( Q )$
• B. $( A ) \rightarrow( P ) ;( B ) \rightarrow( S ) ;( C ) \rightarrow( Q ) ;( D ) \rightarrow( P )$
• C. $( A ) \rightarrow( P ) ;( B ) \rightarrow( Q ) ;( C ) \rightarrow( Q ) ;( D ) \rightarrow( S )$
• D. $( A ) \rightarrow( Q ) ;( B ) \rightarrow( S ) ;( C ) \rightarrow( R ) ;( D ) \rightarrow( P )$

Answer 1

Answer: (A)

(A) ENDEA, $N, O, E, L$ are five different letter, then permutation $=5 !$
(B) If $E$ is in the first and last position then
$\frac{(9-2) !}{2 !}=7 \times 3 \times 5 !=21 \times 5 !$
(C) for first four letters $=\frac{4 !}{2 !}$
for last five letters $=5 ! / 3 !$
Hence $\frac{4 !}{2 !} \times \frac{5 !}{3 !}=2 \times 5 !$
(D) For $A , E$ and $O 5 ! / 3 !$ and for others $4 ! / 2 !$
hence $\frac{5 !}{3 !} \times \frac{4 !}{2 !}=2 \times 5 !$