Step 1: Understanding the Concept:
This question requires calculation and solving of permutations (\(^{n}P_{r}\)) and combinations (\(^{n}C_{r}\)).
Step 2: Key Formula or Approach:
Permutation formula: \(^{n}P_{r} = \frac{n!}{(n-r)!} = n(n-1)(n-2)...(n-r+1)\)
Combination formula: \(^{n}C_{r} = \frac{n!}{r!(n-r)!} = \frac{n(n-1)...(n-r+1)}{r!} \)
Step 3: Detailed Explanation:
(A) \(^{8}P_{3} - ^{10}C_{3}\):
\(^{8}P_{3} = 8 \times 7 \times 6 = 336\)
\(^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120\)
Value = \(336 - 120 = 216\). So, (A) matches (III).
(B) \(^{8}P_{5}\):
\(^{8}P_{5} = 8 \times 7 \times 6 \times 5 \times 4 = 56 \times 120 = 6720\). So, (B) matches (IV).
(C) \(^{n}P_{4} = 360\), find n:
\(n(n-1)(n-2)(n-3) = 360\). We need to find four consecutive integers whose product is 360.
We can test values. Let's try n=6: \(6 \times 5 \times 4 \times 3 = 30 \times 12 = 360\). This is correct.
So, n = 6. (C) matches (I).
(D) \(^{n}C_{2} = 210\), find n:
\(\frac{n(n-1)}{2} = 210\)
\(n(n-1) = 420\). We need two consecutive integers whose product is 420.
We know \(20 \times 20 = 400\). Let's try n=21.
\(21 \times (21-1) = 21 \times 20 = 420\). This is correct.
So, n = 21. (D) matches (II).
The correct matching is: (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Step 4: Final Answer:
The correct option that reflects this matching is (3).