Exercise : 2


1. Out of 15 students studying in a class, 7 are from Maharashtra, 5 are from Karnataka and 3 are from Goa. Four students are to be selected at random. What are the chances that at least one is from Karnataka?
(a) 1213
(b) 1113
(c) 1015
(d) 115
(e) None of these
Ans.b

Total possible ways of selecting 4 students out of 15 students = 15C4 = \(\frac{15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4} = 1365\)

The no. of ways of selecting 4 students in which no student belongs to Karnataka = 10C4

∴ Hence no. of ways of selecting at least one student from Karnataka = 15C410C4 = 1155

∴ Probability = 11551365 = 7791 = 1113

2. 4 boys and 2 girls are to be seated in a row in such a way that the two girls are always together. In how many different ways can they be seated?
(a) 120
(b) 720
(c) 148
(d) 240
(e) None of these
Ans.d

Assume the 2 given students to be together (i.e one).

Now there are five students.

Possible ways of arranging them are = 5! = 120

Now, they (two girls) can arrange themselves in 2! ways.

Hence total ways = 120 × 2 = 240

3. In how many different ways can the letters of the word DETAIL be arranged in such a way that the vowels occupy only the odd positions?
(a) 120
(b) 60
(c) 48
(d) 32
(e) None of these
Ans.e

3 vowels can be arranged in three odd places in 3! ways.

Similarly, 3 consonants can be arranged in three even places in 3! ways.

Hence, the total number of words in which vowels occupy odd positions = 3! × 3! = 6 × 6 = 36 ways.

4. In a box carrying one dozen of oranges, one-third have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one orange out of the three oranges picked up is good?
(a) 155
(b) 5455
(c) 4555
(d) 355
(e) None of these
Ans.b

n(S) = 12C3 = \(\frac{12 \times 11 \times 10}{3 \times 2} = 2 \times 11 \times 10 = 220\)

No. of selection of 3 oranges out of the total 12 oranges

= 12C3 = 2 × 11 × 10 = 220

No. of selection of 3 bad oranges out of the total 4 bad oranges = 4C3 = 4

\n(E) = no. of desired selection of oranges

= 220 – 4 = 216

\ P(E) = \(\frac{n\left ( E \right )}{n\left ( S \right )} = \frac{216}{220} = \frac{54}{55}\)

5. Letters of the word DIRECTOR are arranged in such a way that all the vowels come together. Find out the total number of ways for making such arrangement.
(a) 4320
(b) 2720
(c) 2160
(d) 1120
(e) None of these
Ans.c

Taking all vowels (IEO) as a single letter (since they come together) there are six letters

Hence no. of arrangements = 6!2! × 3! = 2160

[Three vowels can be arranged 3! ways among themselves, hence multiplied with 3!.]
6. A box contains 5 green, 4 yellow and 3 white marbles, 3 marbles are drawn at random. What is the probability that they are not of the same colour?
(a) 1344
(b) 4144
(c) 1355
(d) 5255
(e) None of these
Ans.b

Total no. of ways of drawing 3 marbles

= 12C3 = \(\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220\)

Total no. of ways of drawing marbles, which are of same colour = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15

\Probability of same colour = 15220 = 344

\ Probability of not same colour = 1 - 344 = 4142

7. How many different letter arrangements can be made from the letters of the word RECOVER?
(a) 1210
(b) 5040
(c) 1260
(d) 1200
(e) None of these
Ans.c

Possible arrangements are :

\(\frac{7!}{2! \times 2!} = 1260\) [division by 2 times 2! is because of the repetition of E and R]
8. How many three digit numbers can having only two consecutive digits identical is
(a) 153
(b) 162
(c) 168
(d) 163
(e) None of these
Ans.b

When 0 is the repeated digit like

100, 200, ...., 9 in number

When 0 occurs only once like

110, 220, ....., 9 in number

When 0 does not occur like

112,211, ....., 2 × (8 × 9) = 144 in number.

Hence, total = 9 + 9 + 144 = 162.

9. How many total numbers of seven-digit numbers can be formed having sum of whose digits is even is
(a) 9000000
(b) 4500000
(c) 8100000
(d) 4400000
(e) None of these
Ans.b

Suppose x1 x2 x3 x4 x5 x6 x7 represents a seven digit number. Then x1 takes the value 1, 2, 3, ....., 9 and x2, x3,....., x7 all take values 0, 1, 2, 3, ......, 9.

If we keep x1, x2, ......, x6 fixed, then the sum x1 + x2 + ......+ x6 is either even or odd. Since x7 takes 10 values 0, 1, 2, ....., 9, five of the numbers so formed will have sum of digits even and 5 have sum odd.

Hence the required number of numbers

= 9 . 10 . 10 . 10 . 10 . 10 . 5 = 4500000.

10. How many total numbers of not more than 20 digits that can be formed by using the digits 0, 1, 2, 3, and 4 is
(a) 520
(b) 520 – 1
(c) 520 + 1
(d) 620
(e) None of these
Ans.a

Number of single digit numbers = 5

Number of two digits numbers = 4 × 5

∵ [0 cannot occur at first place and repetition is allowed]

Number of three digits numbers

= 4 × 5 × 5 = 4 × 52

....     ....    ....    ....
....     ....    ....    ....

Number of 20 digits numbers = 4 × 519

∴ Total number of numbers

= 5 + 4.5 + 4.52 + 4.53........4.519

= 5 + \(4 \cdot \frac{5\left ( 5^{19} - 1 \right )}{5 - 1}\) = 5 + 520 - 5 = 520