Permutation and Combinations Questions for IBPS Clerk Set-2
Download important Permutation and Combinations PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Permutation and Combinations for IBPS Clerk Exam.
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Question 1: In a box carrying one dozen of oranges one third have become bad.If 3 oranges taken out from the box random ,what is the probability that at least one orange out of the 3 oranges picked up is good ?
a) 1/55
b) 54/55
c) 45/55
d) 3/55
e) None of these
Question 2: How many such pairs of letters are there in the word FOREHAND each of which have as many letters between them in the word as they have in the English alphabet?
a) None
b) One
c) Two
d) Three
e) More than three
Question 3: A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one
ball is drawn from each bag, and the probability that both are green.
a) 13/70
b) 1/4
c) 6/35
d) 8/35
e) None of these
Question 4: A bag contains 2 red, 3 green and 2 blue balls. 2 balls are to be drawn randomly. What is the probability that the balls drawn contain no blue ball ?
a) 5/7
b) 10/21
c) 2/7
d) 11/21
e) None of these
Question 5: There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls ?
a) $\frac{191}{1547}$
b) $\frac{180}{1547}$
c) $\frac{280}{1547}$
d) $\frac{189}{1547}$
e) None of these
Question 6: In how many different ways can the letters of the word ‘REPLACE’ be arranged ?
a) 2630
b) 5040
c) 1680
d) 2580
e) None of these
Question 7: A bag contains 16 eggs out of which 5 are rotten. The remaining eggs are in good condition. If two eggs are drawn randomly, what is the probability that exactly one of the eggs drawn is rotten ?
a) $\frac{11}{24}$
b) $\frac{13}{24}$
c) $\frac{65}{12}$
d) $\frac{17}{24}$
e) $\frac{7}{12}$
Question 8: A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random. What is the probability that none of the balls drawn are yellow in colour?
a) $\frac{1}{7}$
b) $\frac{3}{7}$
c) $\frac{2}{7}$
d) $\frac{5}{14}$
e) $\frac{9}{14}$
Question 9: Two girls and four boys are to be seated in a row, in such a way that the girls do not sit together. In how many different ways can it be done ?
a) 720
b) 480
c) 360
d) 240
e) None of these
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Question 10: 4 boys and three girls are to be seated in a row in such a way that no two boys sit adjacent to each other. In how many different ways can it be done?
a) 5040
b) 30
c) 144
d) 72
e) None of these
Question 11: In a sample , if a person is picked up randomly, the probability that the person is a smoker is $\frac{3}{5}$, and that of the person being male is $\frac{1}{2}$ .What is the probability that the person is both male as well as a smoker ?
a) $\frac{10}{11}$
b) $\frac{1}{5}$
c) $\frac{3}{5}$
d) Cannot be determined
e) None of these
Question 12: A select group of 4 is to be formed from 8 men and 6 women in such a way that the group must have atleast one woman. In how many different ways can it be done ?
a) 364
b) 1001
c) 728
d) 931
e) None of these
Question 13: Uma has three children, what is the probability that none of the children is a girl ?
a) $\frac{1}{2}$
b) $\frac{1}{16}$
c) $\frac{1}{3}$
d) $\frac{3}{4}$
e) None of these
Question 14: A committee of 3 members is to be selected out of 3 men and 2 women What is the probability that the committee has at least one woman ?
a) $\frac{1}{10}$
b) $\frac{9}{20}$
c) $\frac{9}{10}$
d) $\frac{1}{20}$
e) None of these
Question 15: Out of 5 girls and 3 boys, a total of 4 children are to be randomly selected for a quiz contest. What is the probability that all the four children are girls?
a) $\frac{1}{14}$
b) $\frac{1}{7}$
c) $\frac{5}{17}$
d) $\frac{2}{17}$
e) None of these
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Answers & Solutions:
1) Answer (B)
Total number of oranges in the box = 12
Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$
= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$
Number of oranges which became bad = $\frac{12}{3}=4$
Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$
Number of desired selection of oranges, n(E) = 220 – 4 = 216
$\therefore$ $P(E) = \frac{n(E)}{n(S)}$
= $\frac{216}{220}= \frac{54}{55}$
=> Ans – (B)
2) Answer (C)
Word = FOREHAND
There are 2 pairs of letters which have as many letters between them in the word as they have in the English alphabet
FA and RN
3) Answer (C)
Total balls in bag A = 4 + 6 = 10
Probability that ball is green = $\frac{4}{10}$
Total balls in bag B = 3 + 4 = 7
Probability that ball is green = $\frac{3}{7}$
=> Required probability = $\frac{4}{10} \times \frac{3}{7}$
= $\frac{6}{35}$
4) Answer (B)
Total number of balls = 2 + 3 + 2 = 7
Total number of outcomes = Drawing 2 balls out of 7
= $C^7_2 = \frac{7 \times 6}{1 \times 2} = 21$
Favourable outcomes = Drawing 2 balls out of 5 (so that none is blue)
= $C^5_2 = \frac{5 \times 4}{1 \times 2} = 10$
=> Required probability = $\frac{10}{21}$
5) Answer (C)
Total number of balls in the bag = 8 + 4 + 5 = 17
P(S) = Total possible outcomes
= Selecting 5 balls at random out of 17
=> $P(S) = C^{17}_5 = \frac{17 \times 16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4 \times 5}$
= $6188$
P(E) = Favorable outcomes
= Selecting 2 brown, 1 orange and 2 black balls.
=> $P(E) = C^8_2 \times C^4_1 \times C^5_2$
= $\frac{8 \times 7}{1 \times 2} \times 4 \times \frac{5 \times 4}{1 \times 2}$
= $28 \times 4 \times 10 = 1120$
$\therefore$ Required probability = $\frac{P(E)}{P(S)}$
= $\frac{1120}{6188} = \frac{280}{1547}$
6) Answer (E)
The word ‘REPLACE’ has 2 E,1 A,1 R,1 C,1 L,1 P.
Hence no. of ways in which it can be arranged=$\frac{7!}{2!}$.
=$\frac{5040}{2}$.
=$2520$.
Hence, Option E is correct
7) Answer (A)
Out of the 16 eggs, 5 eggs are rotten and 11 eggs are in good condition.
According to the question, out of the two eggs drawn one is rotten and the other is in good condition.
Hence, required probability = $ \frac{^5C_{1} * ^{11}C_{1}}{^{16}C_{2}} = \frac{5*11}{16*15/2} = \frac{11}{24}$
Hence, option A is the right choice.
8) Answer (B)
Number of non-yellow balls = 4 + 6 = 10
Required Probability = $\frac{^{10}C_2}{^{15}C_2} = \frac{3}{7}$
9) Answer (B)
In order to solve this question, let us find the total number of ways of arranging four boys and two girls and remove the number of arrangements in which both the girls sit together.
Total number of ways of arranging 4 boys and 2 girls is 6! = 720
Total number of ways of arranging them in such a way that both the girls sit together is (6-1)! * 2 = 240
Hence, the answer is 720 – 240 = 480
10) Answer (C)
3 girls can be seated in 3! ways
The required arrangement is B G B G B G B
4 boys can be seated in 4! ways
Number of required ways = $3! \times 4!$ = 144
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11) Answer (D)
Let’s assume the sample size is 100. Let the number of male smokers be x.
Total number of smokers = 3/5 * 100 = 60
Number of men = 1/2 * 100 = 50
Hence, number of male non-smokers is 50-x. Number of female smokers is 60-x and number of female non-smokers is x-10.
Hence, probability of a person picked at random being a smoker and a male = x/100
As we do not know the value of x, we cannot determine the probability. Hence, option D.
12) Answer (D)
The total number of people is 8+6 = 14
An easy way to calculate the number of ways of selecting atleast one woman is to calculate the opposite of it.
The total number of ways of selecting 4 people from a total of 14 is $^{14}C_4 = 1001$
The number of ways of selecting a group of all men (and no women) is $^8C_4 = 70$
Hence, the number of ways of selecting 4 people such that atleast one of them is a woman is $1001-70 = 931$
13) Answer (E)
The number of possible combinations is 2 * 2 * 2 = 8
The probability that none of the children is a girl is 1/8
Option e) is the correct answer.
14) Answer (C)
The number of ways = Two men and one woman + One men and two women
= $\frac{3C2 * 2C1}{5C3} + \frac{3C1 * 2C2}{5C3}
= \frac{6+3}{20}
= 9/20
Hence, the correct option is C.
15) Answer (A)
The total number of ways in which 4 children can be selected is $^8C_4 = 70$
The number of favorable ways in which this can be done is $^5C_4 = 5$
Hence, the required number of ways equals $\frac{5}{70} = \frac{1}{14}$
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