5. (10 points) The sum of the ages of four people, A, B, C, and D, this year is 72 years. A few years ago (at least one year), when A was 22 years old, B was 16 years old. It is also known that when A was 19 years old, C's age was 3 times D's age (at that time, D was at least 1 year old). If the ages of A, B, C, and D are all different, how many possible ages can A have this year?
A. 4
B. 6
C. 8
D. 10
【Analysis】It is known that the sum of the ages of four people this year is 72 years. A few years ago (at least one year), when A was 22 years old, B was 16 years old. When A was 19 years old, B was 13 years old, and C's age was three times D's age (at this time D is at least 1 year old). Since this was at least 4 years ago, the sum of the ages of the four people does not exceed $72-4 \times 4=56$ (years). Let D's age at that time be $x$ years, and C's age be $3x$ years. Then $19+13+x+3x \leqslant 56$, solving this gives $x \leqslant 6$. Knowing that $x \geqslant 1$, D's age has 6 possible scenarios. If the ages of A, B, C, and D are all different, then A's age this year can have 6 possible scenarios.

【Solution】Solution: At least one year ago, when A was 22 years old, B was 16 years old.
When A was 19 years old, B was 13 years old, and C's age was three times D's age (at this time D is at least 1 year old).
Since this was at least 4 years ago, the sum of the ages of the four people does not exceed $72-4 \times 4=56$ (years). Let D's age at that time be $x$ years, and C's age be $3x$ years.

Then $19+13+x+3x \leqslant 56$
$$
\begin{array}{r}
4x+32 \leqslant 56 \\
4x \leqslant 24 \\
x \leqslant 6
\end{array}
$$

Knowing that $x \geqslant 1$, D's age has 6 possible scenarios.
Since the ages of A, B, C, and D are all different, A's age this year can have 6 possible scenarios.
Therefore, the answer is: $B$.