Match the List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (Expression/Statements etc.) | (Values etc.) | ||
| A. | 5P5 − 10C0 | I. | 22 |
| B. | 15C11 − 16C2 | II. | 2520 |
| C. | If nP3 = 9240, then find the value of n. | III. | 0 |
| D. | In how many ways, the letters of the word ‘BANKING’ be arranged? | IV. | 1245 |
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
Each item is worked out and then matched with its value. Two formulas are used:
Arrangements of r things out of n: \(\displaystyle {}^{n}P_{r}=\frac{n!}{(n-r)!}\)
Selections of r things out of n: \(\displaystyle {}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}\)
(A) 5P5 − 10C0 — this matches (III) 0, by elimination: the other three values are taken by (B), (C) and (D). As printed, the expression does not give 0:
The first term is \({}^{5}P_{5}=5!=120\)
The second term is \({}^{10}C_{0}=1\)
The difference is \(120-1=119\), which is not in List-II
(B) 15C11 − 16C2 — this matches (IV) 1245.
Choosing 11 of 15 is the same as leaving out 4: \({}^{15}C_{11}={}^{15}C_{4}\)
The first term works out to \(\displaystyle {}^{15}C_{4}=\frac{15\times14\times13\times12}{24}=1365\)
The second term is \(\displaystyle {}^{16}C_{2}=\frac{16\times15}{2}=120\)
The difference is \(1365-120=1245\)
(C) If nP3 = 9240, then find the value of n. — this matches (I) 22.
\({}^{n}P_{3}=n(n-1)(n-2)\), a product of three consecutive numbers
9240 splits as \(22\times21\times20\), so \(n=22\)
(D) In how many ways, the letters of the word ‘BANKING’ be arranged? — this matches (II) 2520.
BANKING has 7 letters, with N repeated twice
Dividing by \(2!\) removes the repeated arrangements of the two N’s: \(\displaystyle \frac{7!}{2!}=\frac{5040}{2}=2520\)
Note: It seems that there is a mistake in the original NTA paper, which prints item (A) as 5P5 − 10C0, which works out to 119, a value not in List-II. The item was most likely meant to be 5C5 − 10C0 = 1 − 1 = 0. Option (3) is still the only possible answer, because B - IV and C - I appear only in Option (3). One letter, P in place of C, turned 0 into 119; elimination still gets you the marks.