Question

Match List-I with List-II

List-I	List-II
(A) \(^{8}P_{3} - ^{10}C_{3}\)	(I) 6
(B) \(^{8}P_{5}\)	(II) 21
(C) \(^{n}P_{4} = 360,\) then find \(n\).	(III) 216
(D) \(^{n}C_{2} = 210,\) find \(n\).	(IV) 6720

Choose the correct answer from the options given below:

• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
• (A) - (IV), (B) - (I), (C) - (II), (D) - (III)

Hint:

For solving equations like \(^{n}P_{r} = k\) or \(^{n}C_{r} = k\), instead of expanding into complex polynomials, try to estimate and test integer values. Look for consecutive numbers whose product is close to k.

Answer 1

The Correct Option is C

Step 1: Conceptual Understanding:
This problem involves calculations using permutation (\(^{n}P_{r}\)) and combination (\(^{n}C_{r}\)) formulas.
Step 2: Key Formulas:
Permutation: \(^{n}P_{r} = \frac{n!}{(n-r)!} = n(n-1)(n-2)...(n-r+1)\)
Combination: \(^{n}C_{r} = \frac{n!}{r!(n-r)!} = \frac{n(n-1)...(n-r+1)}{r!} \)
Step 3: Detailed Calculations:
(A) \(^{8}P_{3} - ^{10}C_{3}\):
\(^{8}P_{3} = 8 \times 7 \times 6 = 336\)
\(^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\)
Result = \(336 - 120 = 216\). Thus, (A) corresponds to (III).
(B) \(^{8}P_{5}\):
\(^{8}P_{5} = 8 \times 7 \times 6 \times 5 \times 4 = 6720\). Thus, (B) corresponds to (IV).
(C) \(^{n}P_{4} = 360\), determine n:
\(n(n-1)(n-2)(n-3) = 360\). Find four consecutive integers whose product is 360.
Testing n=6: \(6 \times 5 \times 4 \times 3 = 360\). Correct.
Therefore, n = 6. (C) corresponds to (I).
(D) \(^{n}C_{2} = 210\), determine n:
\(\frac{n(n-1)}{2} = 210\)
\(n(n-1) = 420\). Find two consecutive integers whose product is 420.
Approximation: \(20^2 = 400\). Testing n=21.
\(21 \times (21-1) = 21 \times 20 = 420\). Correct.
Therefore, n = 21. (D) corresponds to (II).
The correct matches are: (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Step 4: Final Result:
The option representing this matching is (3).