2015 AMC 10B Problem 16

The assignment is an ordered triple of distinct numbers, so there are $10 \cdot 9 \cdot 8 = 720$ equally likely outcomes.

Let the numbers be $a$ (Al), $b$ (Bill), $c$ (Cal) with $b \mid a$ and $c \mid b$. Since the numbers are distinct, each multiple is a proper multiple, so $c < b < a \le 10$. In particular $c \le 2$ (if $c \ge 3$ then $b \ge 6$ and $a \ge 12$).

Case $c = 1$: choose $b$, then $a$ a larger multiple of $b$ up to $10$.
- $b = 2$: $a \in \{4, 6, 8, 10\}$, $4$ ways.
- $b = 3$: $a \in \{6, 9\}$, $2$ ways.
- $b = 4$: $a = 8$, $1$ way.
- $b = 5$: $a = 10$, $1$ way.
- $b \ge 6$: no $a$.

That is $8$ triples.

Case $c = 2$: $b \in \{4, 6, 8, 10\}$, and $a$ must be a larger multiple of $b$ at most $10$; only $b = 4$, $a = 8$ works. That is $1$ triple.

Total favorable: $8 + 1 = 9$. The probability is $\frac{9}{720} = \frac{1}{80}$.

The answer is $\boxed{\textbf{(C)}\ \frac{1}{80}}$.

A divisibility chain among distinct positive integers is strictly increasing and at least doubles at each step, which caps the smallest element immediately and keeps the enumeration to a handful of lines. Fix the most constrained variable (Cal's, the smallest) and list outward. Keep numerator and denominator in the same model: both count ordered assignments.

• Using 1000 or C(10,3) = 120 as the denominator instead of the 10 9 8 = 720 ordered assignments of distinct numbers.
• Allowing $a = b$ or $b = c$ (every number is a multiple of itself), which inflates the count because the problem requires distinct numbers.