Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \({}^{n}C_{r-1}+{}^{n}C_{r}\) | (I) | \({}^{n}C_{n-r}\) |
| (B) | \({}^{n}C_{r}\) | (II) | \({}^{n+1}C_{r}\) |
| (C) | \({}^{50}C_{r}={}^{50}C_{r+2}\), then r = | (III) | 24 |
| (D) | \({}^{n}P_{3}=9240\), n = ? | (IV) | 22 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
Two results on combinations decide (A), (B) and (C): Pascal’s rule \({}^{n}C_{r-1}+{}^{n}C_{r}={}^{n+1}C_{r}\), and the rule of complements \({}^{n}C_{r}={}^{n}C_{n-r}\), because choosing \(r\) things is the same as leaving out the other \(n-r\).
(A) \({}^{n}C_{r-1}+{}^{n}C_{r}\) — this matches (II) \({}^{n+1}C_{r}\). This is Pascal’s rule.
(B) \({}^{n}C_{r}\) — this matches (I) \({}^{n}C_{n-r}\). This is the rule of complements.
(C) \({}^{50}C_{r}={}^{50}C_{r+2}\), then r = — this matches (III) 24. Since \(r\ne r+2\), equal values need the two lower numbers to add up to 50 (rule of complements): \(r+(r+2)=50\), so \(2r=48\) and \(r=24\).
(D) \({}^{n}P_{3}=9240\), n = ? — this matches (IV) 22. \({}^{n}P_{3}=n(n-1)(n-2)\) is a product of three consecutive numbers, and \(22\times21\times20=9240\), so \(n=22\).