Q.Verify that $^{10}P_{3} = {}^{10}C_{3} \times 3!$, and hence state the general relationship connecting $^{n}P_{r}$ and $^{n}C_{r}$.

[!TLDR] Compute both sides independently and confirm they match; state $^{n}P_{r}={}^{n}C_{r}\times r!$. [!ANSWER] Both sides equal $720$; the …

Left side: $^{10}P_{3} = 10\times9\times8 = 720$. Right side: $^{10}C_{3} = \dfrac{10\times9\times8}{3!} = \dfrac{720}{6} = 120$, so $^{10}C_{3}\times3! = 120\times6 = 720$. Both sides equal $720$, confirming the identity for these particular values of $n$ and $r$. In general, any ordered selection ($^{n}P_{r}$) can be built by first choosing which $r$ objects to use ($^{n}C_{r}$ ways) and then arranging those $r$ objects among …

Compute $^{10}P_{3}$ directly, compute $^{10}C_{3}$ and multiply by $3!$ separately, confirm the two match, then state th …

A common mistake is only verifying the numerical equality without being able to state or explain the underlying general re …