【Solution】Let the number of times A, B, and C are hit be $x$, $y$, and $z$ respectively, then the number of shots fired by the three are $6x$, $5y+1$, and $4z$ respectively. According to the problem, we have the equation: $6x + 5y + 1 + 4z - (x + y + z) = 16$

Simplifying, we get: $5x + 4y + 3z = 15$. First, consider the values of $x$: $x = 3, 1, 0$
1) When $x = 3$, $y = z = 0$; this does not meet the problem's conditions, so it is discarded;
2) When $x = 1$, $y = 1$, $z = 2$;
3) When $x = 0$, $y = 3$, $z = 1$; or 4) $x = 0$, $y = 0$, $z = 5$ (this does not meet the problem's conditions, so it is discarded). There are 2 different situations for the number of times A, B, and C are hit, so the answer is $B$.
6. (10 points) Three people, A, B, and C, are participating in a special real-life $CS$ game, with the following rules: The first shot is fired by B, targeting either A or C. In the subsequent shooting process, if A is hit once, A can shoot 6 bullets at B or C; if B is hit once, B can shoot 5 bullets at A or C; if C is hit once, C can shoot 4 bullets at A or B. At the end of the game, a total of 16 bullets have not hit anyone. How many different scenarios are there for the number of times A, B, and C have been hit?
A. 1
B. 2
C. 3
D. 4