Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \({}^{75}P_{2}-{}^{75}C_{2}\) | (I) | 504 |
| (B) | \({}^{5}P_{5}-{}^{10}C_{3}\) | (II) | 6 |
| (C) | \({}^{16}C_{13}-{}^{8}C_{3}\) | (III) | 2775 |
| (D) | \({}^{n}P_{4}=360\), then find n | (IV) | 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
\({}^{n}P_{r}\) counts arrangements (order matters) and \({}^{n}C_{r}\) counts selections (order does not matter):
\({}^{n}C_{r}={}^{n}P_{r}\div r!\)
\({}^{n}C_{r}={}^{n}C_{n-r}\)
(A) \({}^{75}P_{2}-{}^{75}C_{2}\) — this matches (III) 2775.
\({}^{75}P_{2}=75\times74=5550\)
\({}^{75}C_{2}=5550\div2=2775\)
Difference \(=5550-2775=2775\)
(B) \({}^{5}P_{5}-{}^{10}C_{3}\) — this matches (IV) 0.
\({}^{5}P_{5}=5!=120\)
\({}^{10}C_{3}=\displaystyle\frac{10\times9\times8}{3\times2\times1}=120\)
Difference \(=120-120=0\)
(C) \({}^{16}C_{13}-{}^{8}C_{3}\) — this matches (I) 504.
\({}^{16}C_{13}={}^{16}C_{3}=\displaystyle\frac{16\times15\times14}{6}=560\)
\({}^{8}C_{3}=\displaystyle\frac{8\times7\times6}{6}=56\)
Difference \(=560-56=504\)
(D) \({}^{n}P_{4}=360\), then find n — this matches (II) 6.
\({}^{n}P_{4}=n(n-1)(n-2)(n-3)\), a product of four consecutive numbers
\(360=6\times5\times4\times3\), so \(n=6\)