Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (Expression) | (Value) | ||
| (A) | \(\displaystyle\frac{12!}{10!\,(2!)}\) | (I) | 110 |
| (B) | \({}^{n}C_{2}=210\), find n. | (II) | 136 |
| (C) | \({}^{6}P_{3}-{}^{5}C_{2}\) | (III) | 66 |
| (D) | If \({}^{n}C_{9}={}^{n}C_{8}\), find \({}^{n}C_{15}\) | (IV) | 21 |
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 \({}^{n}P_{r}=\frac{n!}{(n-r)!}\), and the number of combinations (selections) is \({}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}\). Choosing \(r\) things is the same as leaving out \(n-r\), so \({}^{n}C_{r}={}^{n}C_{n-r}\).
(A) \(\displaystyle\frac{12!}{10!\,(2!)}\) — this matches (III) 66.
\(\displaystyle \frac{12\times11\times10!}{10!\times2}=\frac{132}{2}=66\)
(B) \({}^{n}C_{2}=210\), find \(n\). — this matches (IV) 21.
\(\displaystyle \frac{n(n-1)}{2}=210\), so \(n(n-1)=420=21\times20\)
\(n=21\)
(C) \({}^{6}P_{3}-{}^{5}C_{2}\) — this matches (I) 110.
\({}^{6}P_{3}=6\times5\times4=120\)
\(\displaystyle {}^{5}C_{2}=\frac{5\times4}{2}=10\)
\(120-10=110\)
(D) If \({}^{n}C_{9}={}^{n}C_{8}\), find \({}^{n}C_{15}\) — this matches (II) 136.
Two different \(r\) give the same value only when they add up to \(n\): \(n=9+8=17\)
\(\displaystyle {}^{17}C_{15}={}^{17}C_{2}=\frac{17\times16}{2}=136\)