Question 21

1. Mathematically:

   $$\binom{n}{3} = \frac{n(n-1)(n-2)}{6}.$$

2. Analyze each option for large $n$ in 32-bit unsigned integer arithmetic:

   • Option A (p = n * (n-1) * (n-2) / 6):
     Multiplication has left-to-right associativity. The cubic product $n(n-1)(n-2)$ is evaluated first. For $n > 1625$, $n^3 > 2^{32} - 1$, causing 32-bit integer overflow even when $\binom{n}{3}$ itself easily fits in an unsigned integer.
   • Option C (p = n * (n-1) / 3 * (n-2) / 2):
     $n(n-1)$ is not necessarily divisible by 3 (e.g., if $n \equiv 2 \pmod 3$, neither $n$ nor $n-1$ is divisible by 3). In C, integer division / 3 truncates the fraction, yielding an incorrect result.
   • Option D (p = n * (n-1) * (n-2) / 6.0):
     The integer product $n(n-1)(n-2)$ still overflows before the floating-point conversion, and floating-point assignment can cause precision loss.
   • Option B (p = n * (n-1) / 2 * (n-2) / 3):
     Since $n$ and $n-1$ are two consecutive integers, one of them must be even. Therefore, $n(n-1)$ is always divisible by 2, making $n(n-1) / 2$ mathematically exact without truncation. Furthermore, dividing by 2 immediately reduces the intermediate value, substantially delaying overflow. The subsequent multiplication by $(n-2)$ followed by / 3 is also exact because $\binom{n}{3}$ is an integer.

Hence, option B is most likely to compute $p$ correctly.