Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | The number of different words that can be formed with CUSTOM with the condition that the word should begin with M is _____. | (I) | 70 |
| (B) | The number of different ways in which the letters of the word EXTRA can be arranged so that the vowels are never together is _____. | (II) | 45 |
| (C) | There are 10 points in a plane. No three of these points are in a straight line. The total number of straight line that can be formed by joining the two points is ______. | (III) | 72 |
| (D) | The number of ways a committee of 4 people be chosen out of 8 people is _____. | (IV) | 120 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Arranging things in order is a permutation (\(n\) different things in \(n!\) ways); choosing them with order not mattering is a combination, \({}^{n}C_{r}\).
(A) Words from CUSTOM beginning with M — this matches (IV) 120. CUSTOM has 6 different letters; with M fixed first, the other 5 letters fill the remaining places in \(5!=120\) ways.
(B) Arrangements of EXTRA with the vowels never together — this matches (III) 72. It is easier to count the arrangements with the vowels together and subtract them from all arrangements.
All arrangements of the 5 different letters: \(5!=120\)
Vowels E and A together:
Treat EA as one block, so 4 units in \(4!=24\) ways
The block itself in \(2!=2\) ways: \(24\times2=48\)
Vowels never together: \(120-48=72\)
(C) Straight lines through 10 points, no three in a line — this matches (II) 45. Every pair of points gives a different line, so the number of lines is \(\displaystyle {}^{10}C_{2}=\frac{10\times9}{2}=45\).
(D) Committees of 4 out of 8 people — this matches (I) 70. A committee is a selection, so \(\displaystyle {}^{8}C_{4}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70\).