Match each counting problem in List-I with its answer in List-II. List-I (P) The number of arrangements of all letters of BANANA in which no two A's…

(P) Arrange B, N, N first: $\frac{3!}{2!}=3$ ways. The three A's go into distinct gaps among the $4$ gaps: $\binom43=4$. Total $12$: (1).
(Q) A number is divisible by $4$ iff its last two digits are. From digits $1$ to $5$ the valid endings are $12,24,32,44,52$: $5$ choices. The first two digits are free: $5^2=25$. Total $125$: (4).
(R) Without upper bounds: $\binom{12}{2}=66$. Subtract solutions with some $x_i\ge6$: $3\binom{6}{2}=45$ (two variables cannot both be $\ge6$ as $12>10$). Result $21$: (2).
(S) Derangements $D_5=5!\left(1-1+\frac12-\frac16+\frac1{24}-\frac1{120}\right)=44$: (3).
All counts were confirmed by brute-force enumeration. The correct code is P→1, Q→4, R→2, S→3.