Exercise : 3


1. The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat and the terminal digits are even is
(a) 144
(b) 72
(c) 288
(d) 720
(e) None of these
Ans.d

The first and the last (terminal) digits are even and there are three even digits. This arrangement can be done in 3P2 ways. For any one of these arrangements,two even digits are used; and the remaining digits are 5 (4 odd and 1 even) and the four digits in the six digits(leaving out the terminal digits) may be arranged using these 5 digits in 5P4 ways. The required number of numbers is 3P2 × 5P4 = 6 × 120 = 720.

2. Three dice are rolled. The number of possible outcomes in which at least one dice shows 5 is
(a) 215
(b) 36
(c) 125
(d) 91
(e) None of these
Ans.d

Required number of possible outcomes

= Total number of possible outcomes – Number of possible outcomes in which 5 does not appear on any dice. (hence 5 possibilities in each throw)

= 63 – 53 = 216 – 125 = 91

3. The number of ways in which ten candidates A1, A2, ...., A10 can be ranked so that A1 is always above A2 is
(a) 10!2
(b) 10!
(c) 9!
(d) 8!2
(e) None of these
Ans.a

Ten candidates can be ranked in 10! ways. In half of these ways A1 is above A2 and in another half A2 is above A1.

So, required number of ways is 10!2

4. How many total number of ways in which n distinct objects can be put into two different boxes is
(a) n2
(b) 2n
(c) 2n
(d) 3n
(e) None of these
Ans.b

Let the two boxes be B1 and B2. There are two choices for each of the n objects. So, the total number of ways is

2 × 2 ×.....× 2 (n times ) = 2n

5. In how many ways can the letters of the word 'PRAISE' be arranged. So that vowels do not come together?
(a) 720
(b) 576
(c) 440
(d) 144
(e) None of these
Ans.b

Required number of possible outcomes

= Total number of possible outcomes – Number of possible outcomes in which all vowels are together

= 6! – 4 ! × 3! = 576

6. There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?
(a) 144
(b) 180
(c) 192
(d) 360
(e) None of these
Ans.a

Task 1 can not be assigned to either person 1 or 2 i.e.there are 4 options.

Task 2 can be assigned to 3 or 4

So, there are only 2 options for task 2.

So required no. of ways = 2 options for task 2 × 3 options for task 1 × 4 options for task 3 × 3 options for task 4 × 2 options for task 5 × 1 option for task 6.

= 2 × 3 × 4 × 3 × 2 × 1 = 144

7. The number of ways in which one or more balls can be selected out of 10 white, 9 green and 7 blue balls is
(a) 892
(b) 881
(c) 891
(d) 879
(e) None of these
Ans.d

The required number of ways

= (10 + 1)(9 + 1)(7 + 1) - 1 = 879.

8. If 12 persons are seated in a row, the number of ways of selecting 3 persons from them, so that no two of them are seated next to each other is
(a) 85
(b) 100
(c) 120
(d) 240
(e) None of these
Ans.c

The number of ways of selecting 3 persons from 12 people under the given condition :

Number of ways of arranging 3 people among 9 people seated in a row, so that no two of them are consecutive

= Number of ways of choosing 3 places out of the 10[8 in between and 2 extremes]

= 10C3 = \(\frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 5 \times 3 \times 8 = 120\)

9. The number of all possible selections of one or more questions from 10 given questions, each question having one alternative is
(a) 310
(b) 210 – 1
(c) 310 – 1
(d) 210
(e) None of these
Ans.c

Since each question can be selected in 3 ways, by selecting it or by selecting its alternative or by rejecting it. Thus, the total number of ways of dealing with 10 given questions is 310 including a way in which we reject all the questions.

Hence, the number of all possible selections is 310 – 1.

10. A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two of the friends will not attend the party together is
(a) 56
(b) 126
(c) 91
(d) 94
(e) None of these
Ans.c

Number of ways of selecting 5 guests from nine friends = 9C5

Out of these, 7C3 ways are those in which two of the friends occur together [3 more persons to be selected out of remaining 7]

∴ Number of ways, in which two of the friends will not attend the party together = 9C57C3 = 91.