1. The problem involves calculating permutations and combinations for given values $n=6$ and $r=3$. 2. Recall the formulas: - Permutation: $$P(n,r) = \frac{n!}{(n-r)!}$$ - Combination: $$C(n,r) = \frac{n!}{r!(n-r)!}$$ 3. Calculate $P(6,3)$: $$P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!} = \frac{720}{6} = 120$$ 4. Calculate $C(6,3)$: $$C(6,3) = \frac{6!}{3!3!} = \frac{720}{6 \times 6} = \frac{720}{36} = 20$$ 5. The given values in the problem are: - I. $P(6,3) = 120$ (correct) - II. $C(6,3) = 18$ (incorrect, correct value is 20) - III. $P(6,3) = 20$ (incorrect, correct value is 120) 6. Therefore, the correct answers are: - $P(6,3) = 120$ - $C(6,3) = 20$ 7. The rest of the expressions involving factorials and permutations are unrelated to the first question and are not solved here as per instructions. Final answer: $$P(6,3) = 120, \quad C(6,3) = 20$$