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