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.

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

a) $\frac{26}{91}$

b) $\frac{1}{7}$

c) $\frac{199}{364}$

d) $\frac{133}{191}$

e) None of these

1) Answer (E)

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)]/¹⁴C₂ = $\frac{25}{91}$

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

a) $\frac{1}{3}$

b) $\frac{199}{364}$

c) $\frac{165}{364}$

d) $\frac{3}{11}$

e) None of these

2) Answer (B)

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

a) $\frac{4}{7}$

b) $\frac{361}{728}$

c) $\frac{60}{1001}$

d) $\frac{1}{1}$

e) None of these

3) Answer (C)

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

a) $\frac{2}{7}$

b) $\frac{253}{728}$

c) $\frac{10}{21}$

d) $\frac{14}{91}$

e) $\frac{30}{91}$

4) Answer (E)

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 = ¹⁰C₃/¹⁴C₃ = $\frac{30}{91}$

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

a) $\frac{10}{1001}$

b) $\frac{9}{14}$

c) $\frac{17}{364}$

d) $\frac{2}{7}$

e) None of these

5) Answer (A)

Solution:

Out of five blue marbles two can be choosen in ⁵C₂ ways
Out of two red marbles two can be choosen in ²C₂ ways
So the probability of choosing two blue and two red marbles is = (⁵C₂ x ²C₂)/¹⁴C₄ = $\frac{10}{1001}$

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

a) 15

b) 45

c) 60

d) 9

e) None of these

6) Answer (C)

Solution:

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

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

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

a) 12

b) 13

c) 24

d) 52

e) None of these

7) Answer (A)

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

a) $\frac{7}{12}$

b) $\frac{37}{44}$

c) $\frac{5}{12}$

d) $\frac{7}{44}$

e) None of these

8) Answer (B)

Solution:

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

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

a) $\frac{3}{7}$

b) $\frac{1}{2}$

c) $\frac{2}{11}$

d) $\frac{1}{6}$

e) None of these

9) Answer (E)

Solution:

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

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

a) $\frac{7}{44}$

b) $\frac{7}{12}$

c) $\frac{5}{12}$

d) $\frac{1}{44}$

e) None of these

10) Answer (D)

Solution:

Number of ways of selecting 3 green marbles out of 3 green marbles =$^3C_3 = 1$

Number of ways of selecting 3 red marbles out of 4 red marbles = $^4C_3 = 4$

Total ways = 1+4 = 5

Probability =$\frac{5}{^{12}C_3} = 1/44$

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