Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (Expressions) | (Values) | ||
| (A) | \(1/6!+1/7!=x/8!\) Find \(x\) | (I) | 1 |
| (B) | Evaluate: \(\displaystyle\frac{n!}{(n-r)!}, n=6, r=2\) | (II) | 100 |
| (C) | If \({}^{n}C_{9}={}^{n}C_{8}\), find \({}^{n}C_{17}\). | (III) | 64 |
| (D) | \({}^{6}P_{3}-{}^{5}P_{2}\) | (IV) | 30 |
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 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 is \({}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}\). Each List-I item is worked out with these.
(A) \(1/6!+1/7!=x/8!\) Find \(x\) — this matches (III) 64.
\(\displaystyle \frac{8!}{6!}+\frac{8!}{7!}=x\)
\(8!\div6!\) leaves \(8\times7\), and \(8!\div7!\) leaves 8
So \(x=8\times7+8=56+8=64\)
(B) Evaluate: \(\frac{n!}{(n-r)!}, n=6, r=2\) — this matches (IV) 30.
The factors from 4 down to 1 cancel, leaving the top two: \(\displaystyle \frac{6!}{4!}=6\times5=30\)
(C) If \({}^{n}C_{9}={}^{n}C_{8}\), find \({}^{n}C_{17}\). — this matches (I) 1.
Since \({}^{n}C_{r}={}^{n}C_{n-r}\), two equal combinations with different \(r\) have their \(r\) values adding up to \(n\).
\(n=9+8=17\)
\({}^{17}C_{17}=1\), as there is only one way to choose all 17
(D) \({}^{6}P_{3}-{}^{5}P_{2}\) — this matches (II) 100.
3 numbers counting down from 6: \({}^{6}P_{3}=6\times5\times4=120\)
2 numbers counting down from 5: \({}^{5}P_{2}=5\times4=20\)
\(120-20=100\)