Permutation and Combination is an important topic in the Quant section of the SNAP Exam. Quant is a scoring section in SNAP, so it is advised to practice as much as questions from quant. This article provides some of the most important Permutation and Combination Questions for SNAP. One can also download this Free Permutation and Combination Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Permutation and Combination questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practicing.<\/p>\n Download Permutation and Combination Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Instructions<\/b>Instructions: Study the given information carefully to answer the questions that follows.<\/p>\n An urn contains 4 green, 5 blue, 2 red and 3 yellow marbles.<\/p>\n Question 1:\u00a0<\/b>If two marbles are drawn at random, what is the probability that both are red or at least one is red?<\/p>\n a)\u00a0$\\frac{26}{91}$<\/p>\n b)\u00a0$\\frac{1}{7}$<\/p>\n c)\u00a0$\\frac{199}{364}$<\/p>\n d)\u00a0$\\frac{133}{191}$<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Out of two red marbles two can be choosen in 2<\/sup>C2<\/sub> ways Question 2:\u00a0<\/b>If three marbles are drawn at random, what is the probability that at least one is yellow?<\/p>\n a)\u00a0$\\frac{1}{3}$<\/p>\n b)\u00a0$\\frac{199}{364}$<\/p>\n c)\u00a0$\\frac{165}{364}$<\/p>\n d)\u00a0$\\frac{3}{11}$<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n At least one marble is yellow which means that there can be one yellow marble or two yellow marbles or three yellow marbles. Question 3:\u00a0<\/b>If eight marbles are drawn at random, what is the probability that there are equal numbers of marbles of each colour?<\/p>\n a)\u00a0$\\frac{4}{7}$<\/p>\n b)\u00a0$\\frac{361}{728}$<\/p>\n c)\u00a0$\\frac{60}{1001}$<\/p>\n d)\u00a0$\\frac{1}{1}$<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n In order to have equal number of marbles in each colour one has to draw 2 marbles from each colour. Question 4:\u00a0<\/b>If three marbles are drawn at random, what is the probability that none is green?<\/p>\n a)\u00a0$\\frac{2}{7}$<\/p>\n b)\u00a0$\\frac{253}{728}$<\/p>\n c)\u00a0$\\frac{10}{21}$<\/p>\n d)\u00a0$\\frac{14}{91}$<\/p>\n e)\u00a0$\\frac{30}{91}$<\/p>\n 4)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Out of the fourteen marbles four are green in colour Question 5:\u00a0<\/b>If four marbles are drawn at random, what is the probability that two are blue and two are red?<\/p>\n a)\u00a0$\\frac{10}{1001}$<\/p>\n b)\u00a0$\\frac{9}{14}$<\/p>\n c)\u00a0$\\frac{17}{364}$<\/p>\n d)\u00a0$\\frac{2}{7}$<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Out of five blue marbles two can be choosen in 5<\/sup>C2<\/sub> ways Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Instructions<\/b><\/p>\n Study the given information carefully and answer the question that follow: Question 6:\u00a0<\/b>The committee should have 2 trainees and 3 research associates ?<\/p>\n a)\u00a015<\/p>\n b)\u00a045<\/p>\n c)\u00a060<\/p>\n d)\u00a09<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n We have to select 2 trainees out of 3 trainees and 3 resarch associates out of 6 resarch associates.<\/p>\n Required number of ways\u00a0= $^3C_2*^6C_3 = 60$ ways<\/p>\n Question 7:\u00a0<\/b>The committee should have all 4 professors and 1 research associate or all 3 trainees and 2 professors ?<\/p>\n a)\u00a012<\/p>\n b)\u00a013<\/p>\n c)\u00a024<\/p>\n d)\u00a052<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 4 professors and 1 resarch associate can be selected in $^4C_4*^6C_1 = 6$ ways<\/p>\n Similarly 3 trainees and 2 professors would be selected in $^3C_3*^4C_2$ ways = 6<\/p>\n Total ways = 6+6 = 12<\/p>\n Instructions<\/b><\/p>\n Study the given information carefully and answer the question that follow: Question 8:\u00a0<\/b>If the three marbles are picked at random what is the probability that at least one is blue ?<\/p>\n a)\u00a0$\\frac{7}{12}$<\/p>\n b)\u00a0$\\frac{37}{44}$<\/p>\n c)\u00a0$\\frac{5}{12}$<\/p>\n d)\u00a0$\\frac{7}{44}$<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Required probability = 1 – Prob. that no ball is blue = $1- \\frac{^7C_3}{^{12}C_3} = 37\/44$<\/p>\n Question 9:\u00a0<\/b>If two marbles are drawn at random what is the probability that both are red ?<\/p>\n a)\u00a0$\\frac{3}{7}$<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a0$\\frac{2}{11}$<\/p>\n d)\u00a0$\\frac{1}{6}$<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Probability that both the balls are red = $\\frac{^4C_2}{^{12}C_2} = \\frac{1}{11}$<\/p>\n Question 10:\u00a0<\/b>If three marbles are picked at random what is the probability that either all are green or all are red ?<\/p>\n a)\u00a0$\\frac{7}{44}$<\/p>\n b)\u00a0$\\frac{7}{12}$<\/p>\n c)\u00a0$\\frac{5}{12}$<\/p>\n d)\u00a0$\\frac{1}{44}$<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of ways of selecting 3 green marbles out of 3 green marbles\u00a0=$^3C_3 = 1$<\/p>\n Number of ways of selecting 3 red marbles out of 4 red marbles =\u00a0$^4C_3 = 4$<\/p>\n Total ways = 1+4 = 5<\/p>\n Probability =$\\frac{5}{^{12}C_3} = 1\/44$<\/p>\n
\nOut of two red marbles one can be choosen in 2<\/sup>C1<\/sub> ways
\nSum of blue, yellow and green marbles = 12
\nOut of 12 marbles one can be choosen in 12<\/sup>C1<\/sub> ways
\nThe probability that both are red or atleast one is red = [2<\/sup>C2<\/sub> + (2<\/sup>C1<\/sub> x 12<\/sup>C1<\/sub>)]\/14<\/sup>C2<\/sub>\u00a0= $\\frac{25}{91}$<\/p>\n
\nWe know that 0 < probability < 1.
\nSubtractiong the probability of having no yellow marble gives the probability of having at least on yellow marble.
\n1 – (11<\/sup>C3<\/sub>\/14<\/sup>C3<\/sub>) = 1 – $\\frac{165}{364}$ = $\\frac{199}{364}$<\/p>\n
\nOut of four green marbles two can be choosen in 4<\/sup>C2 <\/sub>ways
\nOut of five green marbles two can be choosen in 5<\/sup>C2 <\/sub>ways
\nOut of two red marbles two can be choosen in 2<\/sup>C2<\/sub> ways
\nOut of three yellow marbles 2 can be choosen in 3<\/sup>C2<\/sub> ways
\nSo the probability is = ( 4<\/sup>C2<\/sub> + 5<\/sup>C2 <\/sub>+ 2<\/sup>C2 <\/sub>+ 3<\/sup>C2 <\/sub>)\/(14<\/sup>C8<\/sub>) = $\\frac{60}{1001}$<\/p>\n
\nSubtracting the green marbles from the total marbles we will be left with 10 marbles
\nOut of ten marbles three can be choosen in 10<\/sup>C3 <\/sub>ways
\nProbability of not choosing a choosing marble = 10<\/sup>C3<\/sub>\/14<\/sup>C3\u00a0<\/sub>= $\\frac{30}{91}$<\/p>\n
\nOut of two red marbles two can be choosen in 2<\/sup>C2\u00a0<\/sub>ways
\nSo the probability of choosing two blue and two red marbles is = (5<\/sup>C2\u00a0<\/sub>x\u00a02<\/sup>C2<\/sub>)\/14<\/sup>C4\u00a0<\/sub>= $\\frac{10}{1001}$<\/p>\n
\nA committee of five members is to be formed out of 3 trainees 4 professors and 6 research associates In how many different ways can this be done if____<\/p>\n
\nA basket contains 4 red 5 blue and 3 green marbles.<\/p>\n