Q.(i) Evaluate $^{7}P_{3}$.

Part (ii) is really the same counting problem as part (i), phrased as an application, since arranging 3 chosen books in order out of 7 is exactly what $^{7}P_{3}$ measures.

(i) $^{7}P_{3} = \dfrac{7!}{(7-3)!} = \dfrac{7!}{4!}$. Since $7! = 7\times6\times5\times4!$, this simplifies to $7\times6\times5 = 210$.

(ii) Selecting 3 books from 7 and arranging them in a definite order on the shelf is precisely a permutation of 7 distinct things taken 3 at a time — order matters here because a different arrangement of the same 3 books is a genuinely different outcome on the shelf. So the answer is $^{7}P_{3}=210$, the identical value as part (i). …

Using the combination formula $^{7}C_{3}=35$ instead of $^{7}P_{3}=210$ for part (ii) — since the books are arranged in order on the shelf, the …