ENDEANOEL
(A)→(P)
ENDEA, N, O, E, L are five different letters, then permutation is = 5!.
(B)→(S)
If E is at the first and at the last position, we get
(9−2)!/2! = 7 × 3 × 5! = 21 × 5!
(C)→(Q)
The arrangements of last five letters:
5!/2! = 20
Therefore, the number of permutation is
12 × 20 = 240 = 2 × 5!
(D)→(Q)
The arrangements of O, E and A are
5!/3! and that of other letters are
4!/2!. Therefore, the number of permutation is
5!/3!×4!/2! = 2 × 5!