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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The number of permutations (arrangements) of \(r\) things out of \(n\) is \({}^{n}P_{r}=\frac{n!}{(n-r)!}\), the product of \(r\) numbers counting down from \(n\). The number of combinations (selections) is \({}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}\). Each List-I item is worked out with these.
(A) \({}^{8}P_{3}-{}^{10}C_{3}\) — this matches (III) 216.
\({}^{8}P_{3}=8\times7\times6=336\)
\(\displaystyle {}^{10}C_{3}=\frac{10\times9\times8}{3\times2\times1}=120\)
\(336-120=216\)
(B) \({}^{8}P_{5}\) — this matches (IV) 6720.
\({}^{8}P_{5}=8\times7\times6\times5\times4=6720\)
(C) \({}^{n}P_{4}=360\), then find n. — this matches (I) 6.
\({}^{n}P_{4}=n(n-1)(n-2)(n-3)\), four numbers counting down from \(n\)
\(6\times5\times4\times3=360\), so \(n=6\)
(D) \({}^{n}C_{2}=210\), find n. — this matches (II) 21.
\(\displaystyle {}^{n}C_{2}=\frac{n(n-1)}{2}=210\)
\(n(n-1)=420=21\times20\), so \(n=21\)