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# SNAP Permutation and Combination Questions PDF

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, which is one of the best sources for practicing.

InstructionsInstructions: Study the given information carefully to answer the questions that follows.

An urn contains 4 green, 5 blue, 2 red and 3 yellow marbles.

Question 1:Â If two marbles are drawn at random, what is the probability that both are red or at least one is red?

a)Â $\frac{26}{91}$

b)Â $\frac{1}{7}$

c)Â $\frac{199}{364}$

d)Â $\frac{133}{191}$

e)Â None of these

Solution:

Out of two red marbles two can be choosen in 2C2 ways
Out of two red marbles one can be choosen in 2C1 ways
Sum of blue, yellow and green marbles = 12
Out of 12 marbles one can be choosen in 12C1 ways
The probability that both are red or atleast one is red = [2C2 + (2C1 x 12C1)]/14C2Â = $\frac{25}{91}$

Question 2:Â If three marbles are drawn at random, what is the probability that at least one is yellow?

a)Â $\frac{1}{3}$

b)Â $\frac{199}{364}$

c)Â $\frac{165}{364}$

d)Â $\frac{3}{11}$

e)Â None of these

Solution:

At least one marble is yellow which means that there can be one yellow marble or two yellow marbles or three yellow marbles.
We know that 0 < probability < 1.
Subtractiong the probability of having no yellow marble gives the probability of having at least on yellow marble.
1 – (11C3/14C3) = 1 – $\frac{165}{364}$ = $\frac{199}{364}$

Question 3:Â If eight marbles are drawn at random, what is the probability that there are equal numbers of marbles of each colour?

a)Â $\frac{4}{7}$

b)Â $\frac{361}{728}$

c)Â $\frac{60}{1001}$

d)Â $\frac{1}{1}$

e)Â None of these

Solution:

In order to have equal number of marbles in each colour one has to draw 2 marbles from each colour.
Out of four green marbles two can be choosen in 4C2 ways
Out of five green marbles two can be choosen in 5C2 ways
Out of two red marbles two can be choosen in 2C2 ways
Out of three yellow marbles 2 can be choosen in 3C2 ways
So the probability is = ( 4C2 + 5C2 + 2C2 + 3C2 )/(14C8) = $\frac{60}{1001}$

Question 4:Â If three marbles are drawn at random, what is the probability that none is green?

a)Â $\frac{2}{7}$

b)Â $\frac{253}{728}$

c)Â $\frac{10}{21}$

d)Â $\frac{14}{91}$

e)Â $\frac{30}{91}$

Solution:

Out of the fourteen marbles four are green in colour
Subtracting the green marbles from the total marbles we will be left with 10 marbles
Out of ten marbles three can be choosen in 10C3 ways
Probability of not choosing a choosing marble = 10C3/14C3Â = $\frac{30}{91}$

Question 5:Â If four marbles are drawn at random, what is the probability that two are blue and two are red?

a)Â $\frac{10}{1001}$

b)Â $\frac{9}{14}$

c)Â $\frac{17}{364}$

d)Â $\frac{2}{7}$

e)Â None of these

Solution:

Out of five blue marbles two can be choosen in 5C2 ways
Out of two red marbles two can be choosen in 2C2Â ways
So the probability of choosing two blue and two red marbles is = (5C2Â xÂ 2C2)/14C4Â = $\frac{10}{1001}$

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Instructions

Study the given information carefully and answer the question that follow:
A 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____

Question 6:Â The committee should have 2 trainees and 3 research associates ?

a)Â 15

b)Â 45

c)Â 60

d)Â 9

e)Â None of these

Solution:

We have to select 2 trainees out of 3 trainees and 3 resarch associates out of 6 resarch associates.

Required number of waysÂ = $^3C_2*^6C_3 = 60$ ways

Question 7:Â The committee should have all 4 professors and 1 research associate or all 3 trainees and 2 professors ?

a)Â 12

b)Â 13

c)Â 24

d)Â 52

e)Â None of these

Solution:

4 professors and 1 resarch associate can be selected in $^4C_4*^6C_1 = 6$ ways

Similarly 3 trainees and 2 professors would be selected in $^3C_3*^4C_2$ ways = 6

Total ways = 6+6 = 12

Instructions

Study the given information carefully and answer the question that follow:
A basket contains 4 red 5 blue and 3 green marbles.

Question 8:Â If the three marbles are picked at random what is the probability that at least one is blue ?

a)Â $\frac{7}{12}$

b)Â $\frac{37}{44}$

c)Â $\frac{5}{12}$

d)Â $\frac{7}{44}$

e)Â None of these

Solution:

Required probability = 1 – Prob. that no ball is blue = $1- \frac{^7C_3}{^{12}C_3} = 37/44$

Question 9:Â If two marbles are drawn at random what is the probability that both are red ?

a)Â $\frac{3}{7}$

b)Â $\frac{1}{2}$

c)Â $\frac{2}{11}$

d)Â $\frac{1}{6}$

e)Â None of these

Solution:

Probability that both the balls are red = $\frac{^4C_2}{^{12}C_2} = \frac{1}{11}$

Question 10:Â If three marbles are picked at random what is the probability that either all are green or all are red ?

a)Â $\frac{7}{44}$

b)Â $\frac{7}{12}$

c)Â $\frac{5}{12}$

d)Â $\frac{1}{44}$

e)Â None of these

Number of ways of selecting 3 green marbles out of 3 green marblesÂ =$^3C_3 = 1$
Number of ways of selecting 3 red marbles out of 4 red marbles =Â $^4C_3 = 4$
Probability =$\frac{5}{^{12}C_3} = 1/44$