Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (Expressions) | (Values) | ||
| (A) | \({}^{n}P_{4}=360\), then find n | (I) | 1155 |
| (B) | If \({}^{15}C_{3r}={}^{15}C_{r+3}\), then find r | (II) | 56 |
| (C) | Find \({}^{15}C_{11}-{}^{15}C_{4}\). | (III) | 6 |
| (D) | Find \({}^{8}P_{2}\) | (IV) | 3 |
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 arrangements of \(r\) things out of \(n\) is \({}^{n}P_{r}=\frac{n!}{(n-r)!}\), and the number of selections is \({}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}\). Selections have the property \({}^{n}C_{r}={}^{n}C_{n-r}\): choosing \(r\) things is the same as leaving out the other \(n-r\).
(A) \({}^{n}P_{4}=360\), then find \(n\) — this matches (III) 6.
\({}^{n}P_{4}=n(n-1)(n-2)(n-3)\)
\(6\times5\times4\times3=360\), so \(n=6\)
(B) If \({}^{15}C_{3r}={}^{15}C_{r+3}\), then find \(r\) — this matches (IV) 3. Two equal selections from 15 have either equal lower numbers or lower numbers adding up to 15.
\(3r=r+3\) gives \(r=\frac{3}{2}\), which is not a whole number
So \(3r+(r+3)=15\), giving \(4r=12\) and \(r=3\)
(C) Find \({}^{15}C_{11}-{}^{15}C_{4}\) — this matches (I) 1155, by elimination.
By the property above, \({}^{15}C_{11}={}^{15}C_{15-11}={}^{15}C_{4}=1365\)
So the printed expression is \(1365-1365=0\)
The printed entry 1155 does not actually match it; (C) takes (I) only because (II), (III) and (IV) belong to the other items.
(D) Find \({}^{8}P_{2}\) — this matches (II) 56.
\({}^{8}P_{2}=8\times7=56\)
Note: It seems that there is a mistake in the original NTA paper: item (C), \({}^{15}C_{11}-{}^{15}C_{4}\), equals 0 because \({}^{15}C_{11}={}^{15}C_{4}\), and does not match any value in List-II. The other three items match their values, so Option (3) is still the only possible answer. NTA did not drop the question, so its key still decides the marks: the maths is on our side, the marks are on NTA’s.