AP Intermediate 1st Year Mathematics — practice by topic

Factorial definition: $n! = n \times (n-1) \times (n-2) \times \ldots \times 1$. $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. Quick reference: $4! = 24$, $5! = 120$, $6! = 720$ (option D is $6!$, not $5!$). Factorials grow fast.

Arrangements of $n$ distinct objects in a row: $n$ choices for the first position then $n-1$ for the second, $n-2$ for the third, etc. By the multiplication principle: $n \times (n-1) \times \ldots \times 1 = n!$ For 5 books on a shelf: $5! = 120$ arrangements. For 10 students in a line: $10! = 3,628,800$.

We need to arrange 4 digits chosen from 5 distinct digits with no repetition. That's $^5 P_4$: $^5 P_4 = \dfrac{5!}{(5-4)!} = \dfrac{120}{1} = 120$. Alternative count by positions: 5 choices for the thousands digit then 4 for hundreds, then 3 for tens, then 2 for units. $5 \times 4 \times 3 \times 2 = 120$ ✓. Option B (625) would be allowed repetition: $5^4 = 625$. Option C is just $4!$ ignoring the choice of digits to use.