Download important Permutation and Combination Questions PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Permutation and Combination Question and Answers for IBPS Clerk Exam.<\/p>\n
Download Permutation\u00a0 Combination Questions For IBPS Clerk PDF<\/a><\/p>\n 105 IBPS Clerk for just Rs. 199<\/a><\/p>\n Take Free IBPS Clerk Mock Test<\/a><\/p>\n Download IBPS Clerk Previous papers PDF<\/a><\/p>\n Go to Free Banking Study Material<\/a> (15,000 Solved Questions)<\/p>\n Question 1:\u00a0<\/b>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 ?<\/p>\n a)\u00a01\/55<\/p>\n b)\u00a054\/55<\/p>\n c)\u00a045\/55<\/p>\n d)\u00a03\/55<\/p>\n e)\u00a0None of these<\/p>\n Question 2:\u00a0<\/b>Letters of the word DIRECTOR are arranged in such a way that all the vowel come together .Find the No of ways making such arrangement?<\/p>\n a)\u00a04320<\/p>\n b)\u00a0720<\/p>\n c)\u00a02160<\/p>\n d)\u00a0120<\/p>\n e)\u00a0None of these<\/p>\n Question 3:\u00a0<\/b>Certain number of pieces of an article are to be packed in boxes, such that each box contains 145 pieces. If after packing them in 32 boxes 25 pieces are left out, what was the number of pieces to be packed ?<\/p>\n a)\u00a04566<\/p>\n b)\u00a04655<\/p>\n c)\u00a04465<\/p>\n d)\u00a04640<\/p>\n e)\u00a0None of these<\/p>\n IBPS Clerk Online Mock Test<\/a><\/p>\n Question 4:\u00a0<\/b>A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one a)\u00a013\/70<\/p>\n b)\u00a01\/4<\/p>\n c)\u00a06\/35<\/p>\n d)\u00a08\/35<\/p>\n e)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>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 ?<\/p>\n a)\u00a05\/7<\/p>\n b)\u00a010\/21<\/p>\n c)\u00a02\/7<\/p>\n d)\u00a011\/21<\/p>\n e)\u00a0None of these<\/p>\n Question 6:\u00a0<\/b>In how many different ways can the letters of the word DRASTIC be arranged in such a way that the vowels always come together ?<\/p>\n a)\u00a0720<\/p>\n b)\u00a0360<\/p>\n c)\u00a01440<\/p>\n d)\u00a0540<\/p>\n e)\u00a0None of these<\/p>\n IBPS Clerk Previous Papers<\/a><\/p>\n Question 7:\u00a0<\/b>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 ?<\/p>\n a)\u00a0$\\frac{191}{1547}$<\/p>\n b)\u00a0$\\frac{180}{1547}$<\/p>\n c)\u00a0$\\frac{280}{1547}$<\/p>\n d)\u00a0$\\frac{189}{1547}$<\/p>\n e)\u00a0None of these<\/p>\n Question 8:\u00a0<\/b>A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked at random. What is the probability that the ball drawn is white ?<\/p>\n a)\u00a0$\\frac{7}{11}$<\/p>\n b)\u00a0$\\frac{7}{30}$<\/p>\n c)\u00a0$\\frac{5}{11}$<\/p>\n d)\u00a0$\\frac{7}{15}$<\/p>\n e)\u00a0$\\frac{8}{15}$<\/p>\n Question 9:\u00a0<\/b>A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue in colour ?<\/p>\n a)\u00a0$\\frac{20}{91}$<\/p>\n b)\u00a0$\\frac{10}{91}$<\/p>\n c)\u00a0$\\frac{15}{91}$<\/p>\n d)\u00a0$\\frac{5}{91}$<\/p>\n e)\u00a0$\\frac{25}{91}$<\/p>\n Question 10:\u00a0<\/b>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 ?<\/p>\n a)\u00a0$\\frac{11}{24}$<\/p>\n b)\u00a0$\\frac{13}{24}$<\/p>\n c)\u00a0$\\frac{65}{12}$<\/p>\n d)\u00a0$\\frac{17}{24}$<\/p>\n e)\u00a0$\\frac{7}{12}$<\/p>\n IBPS Clerk Important Questions PDF<\/a><\/p>\n Free Banking Study Material (15,000 Solved Questions)<\/a><\/p>\n Question 11:\u00a0<\/b>A bag contains 5 red balls, 7 yellow balls and 3 pink balls. If two balls are drawn at random from the bag, one after another, what is the probability that the first ball is red and the second ball is yellow ?<\/p>\n a)\u00a0$\\frac{5}{12}$<\/p>\n b)\u00a0$\\frac{3}{8}$<\/p>\n c)\u00a0$\\frac{1}{4}$<\/p>\n d)\u00a0$\\frac{1}{8}$<\/p>\n e)\u00a0$\\frac{1}{6}$<\/p>\n Question 12:\u00a0<\/b>A committee of 6 members is to be selected from a group of 8 men and 6 women in such as way that at least 3 men are there in the committee. In how many different ways can it be done ?<\/p>\n a)\u00a02506<\/p>\n b)\u00a02534<\/p>\n c)\u00a01120<\/p>\n d)\u00a01050<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>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?<\/p>\n a)\u00a05040<\/p>\n b)\u00a030<\/p>\n c)\u00a0144<\/p>\n d)\u00a072<\/p>\n e)\u00a0None of these<\/p>\n Question 14:\u00a0<\/b>A committee of 3 members is to be formed out of 3 men and 4 women. In how many different ways can it be done so that at least one member is a woman?<\/p>\n a)\u00a034<\/p>\n b)\u00a012<\/p>\n c)\u00a030<\/p>\n d)\u00a036<\/p>\n e)\u00a0None of these<\/p>\n Question 15:\u00a0<\/b>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 ?<\/p>\n a)\u00a0$\\frac{10}{11}$<\/p>\n b)\u00a0$\\frac{1}{5}$<\/p>\n c)\u00a0$\\frac{3}{5}$<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n Daily Free Banking Online Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Total number of oranges in the box = 12<\/p>\n Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$<\/p>\n = $\\frac{12 \\times 11 \\times 10}{1 \\times 2 \\times 3} = 220$<\/p>\n Number of oranges which became bad = $\\frac{12}{3}=4$<\/p>\n Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$<\/p>\n Number of desired selection of oranges, n(E) = 220 – 4 = 216<\/p>\n $\\therefore$ $P(E) = \\frac{n(E)}{n(S)}$<\/p>\n = $\\frac{216}{220}= \\frac{54}{55}$<\/p>\n => Ans – (B)<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Word – DIRECTOR<\/p>\n So \u201cI,E,O\u201d are there are 3! ways to arrange the vowels<\/p>\n Now \u201cD,R,C,T,R\u201d are the remaining alphabets ,<\/p>\n Condition is that the vowels should always be together so we can assume the vowels as a single alphabet\/unit say \u201cX\u201d (\u2018X\u2019=\u2019I,E,O\u2019) so now we have a new word – \u201cD,R,C,T,R,X\u201d<\/p>\n Possible arrangements for this word = 6!<\/p>\n Thus total number of ways to rearrange DIRECTOR with vowels grouped together = (Possible arrangements of \u2018DRCTRX\u2019) $\\times$ (Possible arrangements of vowels)<\/p>\n = 6! $\\times$ 3! = $720 \\times 6 = 4320$<\/p>\n => Ans – (A)<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n No. of pieces<\/p>\n = 32 * 145 + 25<\/p>\n = 4640 + 25 = 4665<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Total balls in bag A = 4 + 6 = 10<\/p>\n Probability that ball is green = $\\frac{4}{10}$<\/p>\n Total balls in bag B = 3 + 4 = 7<\/p>\n Probability that ball is green = $\\frac{3}{7}$<\/p>\n => Required probability = $\\frac{4}{10} \\times \\frac{3}{7}$<\/p>\n = $\\frac{6}{35}$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Total number of balls = 2 + 3 + 2 = 7<\/p>\n Total number of outcomes = Drawing 2 balls out of 7<\/p>\n = $C^7_2 = \\frac{7 \\times 6}{1 \\times 2} = 21$<\/p>\n Favourable outcomes = Drawing 2 balls out of 5 (so that none is blue)<\/p>\n =\u00a0$C^5_2 = \\frac{5 \\times 4}{1 \\times 2} = 10$<\/p>\n => Required probability = $\\frac{10}{21}$<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n There are 7 letters in the word ‘DRASTIC’ including 2 vowels (A,I) and 5 consonants (D,R,S,T,C).<\/p>\n Considering the two vowels as 1 letter, we have 6 letters which can be arranged in 6! ways<\/p>\n But \u00a0corresponding to each way of these arrangements, the vowels can be put together in 2! ways.<\/p>\n Hence, required number of words = $6! \\times 2!$<\/p>\n = 720 * 2 = 1440<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Total number of balls in the bag = 8 + 4 + 5 = 17<\/p>\n P(S) = Total possible outcomes<\/p>\n = Selecting 5 balls at random out of 17<\/p>\n => $P(S) = C^{17}_5 = \\frac{17 \\times 16 \\times 15 \\times 14 \\times 13}{1 \\times 2 \\times 3 \\times 4 \\times 5}$<\/p>\n = $6188$<\/p>\n P(E) = Favorable outcomes<\/p>\n = Selecting 2 brown, 1 orange and 2 black balls.<\/p>\n => $P(E) = C^8_2 \\times C^4_1 \\times C^5_2$<\/p>\n = $\\frac{8 \\times 7}{1 \\times 2} \\times 4 \\times \\frac{5 \\times 4}{1 \\times 2}$<\/p>\n = $28 \\times 4 \\times 10 = 1120$<\/p>\n $\\therefore$ Required probability = $\\frac{P(E)}{P(S)}$<\/p>\n = $\\frac{1120}{6188} = \\frac{280}{1547}$<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Probability of choosing bag 1 = (1\/2) 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Probability of drawing blue ball in first attempt = 5\/15<\/p>\n Probability of drawing two green balls in the next two attempts = (6\/14)(5\/13) Probability of drawing green ball in first attempt = 6\/15 Probability of drawing two green balls in first two attempts = (6\/15)(5\/14) Probability of drawing 2 red balls and 1 green ball= 150\/2730 + 150\/2730 + 150\/2730 = 3(150\/2730) = 150\/910 = 15\/91<\/span><\/span> 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Out of the 16 eggs, 5 eggs are rotten and 11 eggs are in good condition.<\/p>\n According to the question, out of the two eggs drawn one is rotten and the other is in good condition.<\/p>\n Hence, required probability = $ \\frac{^5C_{1} * ^{11}C_{1}}{^{16}C_{2}} = \\frac{5*11}{16*15\/2} = \\frac{11}{24}$<\/p>\n Hence, option A is the right choice.<\/p>\n 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Required probability = $\\frac{5}{15} \\times \\frac{7}{14} = \\frac{1}{6}$<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n There are a total of 8 men and 6 women.<\/p>\n Number of ways in which the committee has exactly 3 men is $^8C_3 \\times ^6C_3 = 1120$ Hence, the total is $1120 + 1050 + 336 + 28 = 2534$<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 3 girls can be seated in 3! ways 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Number of ways of selecting any 3 members = $^7C_3 = 35$ 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let’s assume the sample size is 100. Let the number of male smokers be x.<\/p>\n Total number of smokers = 3\/5 * 100 = 60<\/p>\n Number of men = 1\/2 * 100 = 50<\/p>\n 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.<\/p>\n Hence, probability of a person picked at random being a smoker and a male = x\/100<\/p>\n As we do not know the value of x, we cannot determine the probability. Hence, option D.<\/p>\n Highly Rated Free Preparation App for Banking Exams<\/a><\/p>\n
\nball is drawn from each bag, and the probability that both are green.<\/p>\n
\nProbability of choosing bag 2 = (1\/2)
\nProbability of choosing white ball from bag 1 = 3\/5
\nProbability of choosing white ball from bag 2 = 2\/6
\nProbability of choosing bag 1 and white ball from it = (1\/2)(3\/5) = 3\/10
\nProbability of choosing bag 2 and white ball from it = (1\/2)(2\/6) = 2\/12
\nProbability of choosing a bag and drawing a white ball = (3\/10) + (2\/12) = (28\/60) = (7\/15)
\nOption D is the correct answer.<\/span><\/span><\/p>\n
\nProbability of drawing 2 green and 1 blue ball = (5\/15)(6\/14)(5\/13) = 150\/2730<\/p>\n
\nProbability of drawing blue ball in the next attempt = (5\/14)
\nProbability of drawing green ball in the next attempt = (5\/13)
\nProbability of drawing 2 red and 1 green ball = (6\/15)(5\/14)(5\/13) = 150\/2730<\/p>\n
\nProbability of drawing blue ball in the next attempt =(5\/13)
\nProbability of drawing 2 red and 1 green ball = (6\/15)(5\/14)(5\/13) = 150\/2730<\/p>\n
\nOption C is the correct answer<\/span><\/p>\n
\nNumber of ways in which the committee has exactly 4 men is $^8C_4 \\times ^6C_2 = 1050$
\nNumber of ways in which the committee has exactly 5 men is $^8C_5 \\times ^6C_1 = 336$
\nNumber of ways in which the committee has exactly 6 men is $^8C_6 \\times ^6C_0 = 28$<\/p>\n
\nThe required arrangement is B G B G B G B
\n4 boys can be seated in 4! ways
\nNumber of required ways = $3! \\times 4!$ = 144<\/p>\n
\nNumber of ways of selecting only men = $^3C_3 = 1$
\nNumber of ways of selecting such that at least one member is a woman = 35 – 1 = 34<\/p>\n