Q.Evaluate:

A factorial multiplies every whole number down to 1, and $0!$ is fixed at $1$ by definition, not by computation.

1. $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
2. $0! + 1! = 1 + 1 = 2$.
   ✓Final answer

   (i) $120$ (ii) $2$

(i) $5!$

By the definition of factorial, multiply every whole number from $5$ down to $1$:

Step by step: $5\times4=20$, $20\times3=60$, $60\times2=120$, $120\times1=120$.

(ii) $0! + 1!$

$0!$ is defined (an axiom, not derived) to equal $1$. $1! = 1$ trivially (a single number "multiplied" by nothing else).

Treating $0!$ as $0$ instead of the defined value $1$ is the single most common slip in this chapter — it silently breaks every $^{n}P_{r}$/$^{n}C_{r}$ boundary case built on it (e.g. $^{n}P_{0}$ would wrongly come out undefined or zero). Also watch for stopping the multiplication early, e.g. writing $5! = 5\times4\times3\times2 = 120$ but forgetting the final $\times1$ is still needed conceptually (it doesn't change the value here, but skipping it as a habit causes errors in other factorial ratios).