MCQ: Permutations and Combinations - 2 - SSC CGL MCQ

# MCQ: Permutations and Combinations - 2 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Permutations and Combinations - 2

MCQ: Permutations and Combinations - 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Permutations and Combinations - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Permutations and Combinations - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Permutations and Combinations - 2 below.
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MCQ: Permutations and Combinations - 2 - Question 1

### How many different words can be formed using all the letters of the word ALLAHABAD? (a) When vowels occupy the even positions. (b) Both L do not occur together.

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 1

ALLAHABAD = 9 letters. Out of these 9 letters there is 4 A's and 2 L's are there.
So, permutations
(a) There are 4 vowels and all are alike i.e. 4A's.

These even places can be occupied by 4 vowels. In
In other five places 5 other letter can be occupied of which two are alike i.e. 2L's.
Number of ways = 5!/2! Ways.
Hence, total number of ways in which vowels occupy the even places  = 60 ways.

(b) Taking both L's together and treating them as one letter we have 8 letters out of which A repeats 4 times and others are distinct.
These 8 letters can be arranged in = 8!/4! Ways.
Also two L can be arranged themselves in 2! ways.
So, Total no. of ways in which L are together = 1680 × 2 = 3360 ways.
Now,
Total arrangement in which L never occur together,
= Total arrangement - Total no. of ways in which L occur together.
= 7560 - 3360
= 4200 ways

MCQ: Permutations and Combinations - 2 - Question 2

### In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 2

Let the Arrangement be,
B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways
Required number of ways,
= 4! × 3!
= 144

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MCQ: Permutations and Combinations - 2 - Question 3

### In how many ways 2 students can be chosen from the class of 20 students?

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 3

Number of ways

MCQ: Permutations and Combinations - 2 - Question 4

A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 4

There 10 questions in part A out of which 8 question can be chosen as = 10C8
Similarly, 5 questions can be chosen from 10 questions of Part B as = 10C5
Hence, total number of ways,

MCQ: Permutations and Combinations - 2 - Question 5

How many Permutations of the letters of the word APPLE are there?

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 5

APPLE = 5 letters.
But two letters PP is of same kind.
Thus, required permutations,

MCQ: Permutations and Combinations - 2 - Question 6

In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 6

No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!

MCQ: Permutations and Combinations - 2 - Question 7

A two member committee comprising of one male and one female member is to be constitute out of five males and three females. Amongst the females. Ms. A refuses to be a member of the committee in which Mr. B is taken as the member. In how many different ways can the committee be constituted ?

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 7

5C1 × 3C1 - 1
= 15 - 1
= 14

MCQ: Permutations and Combinations - 2 - Question 8

Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 8

There are 6 candidates and a voter has to vote for any two of them.
So, the required number of ways is,

MCQ: Permutations and Combinations - 2 - Question 9

Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 9

A triangle needs 3 points.
And polygon of 8 sides has 8 angular points.
Hence, number of triangle formed,

MCQ: Permutations and Combinations - 2 - Question 10

There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices.

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 10

The number of triangle can be formed by 10 points = 10C3
Similarly, the number of triangle can be formed by 4 points when no one is collinear = 4C3
In the question, given 4 points are collinear, Thus, required number of triangle can be formed,
10C3 - 4C3
= 120 - 4
= 116

MCQ: Permutations and Combinations - 2 - Question 11

The number of positive integers which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 without repetition.

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 11

One digit positive numbers = 5
Two digit positive numbers = 25
Three digit positive numbers = 100
4 digit positive numbers = 300
5 digit positive numbers = 600
Six digit positive numbers = 600
Total positive numbers,
= 5 + 25 + 100 + 300 + 600 + 600
= 1630

MCQ: Permutations and Combinations - 2 - Question 12

A letter lock consists of 4 rings, each ring contains 9 non-zero digits. This lock can be opened by setting four digit code with the proper combination of each of the 4 rings. Maximum how many codes can be formed to open the lock ?

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 12

There are 9 non-zero digits to arrange themselves at 4 different position. Each letter can be arrange at different position in 9 different ways.
So, required number of ways,
= 9 × 9 × 9 × 9
= 94

MCQ: Permutations and Combinations - 2 - Question 13

How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 13

Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4
Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways
Hence, the 4 prizes can be distributed in 24 ways

MCQ: Permutations and Combinations - 2 - Question 14

In a party every person shakes hands with every other person. If there are 105 hands shakes, find the number of person in the party.

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 14

Let n be the number of persons in the party
Number of hands shake = 105
Total number of hands shake is given by nC2
Now,
According to the question,

But, we cannot take negative value of n
So, n = 15
i.e. number of persons in the party = 15

MCQ: Permutations and Combinations - 2 - Question 15

In the next World cup of cricket there will be 12 teams, divided equally in two groups. Teams of each group will play a match against each other. From each group 3 top teams will qualify for the next round. In this round each team will play against each others once. Four top teams of this round will qualify for the semifinal round, where they play the best of three matches. The Minimum number of matches in the next World cup will be:

Detailed Solution for MCQ: Permutations and Combinations - 2 - Question 15

The number of matches in first round,
6C2 +6C2
Number of matches in next round,
6C2
Number of matches in semifinals,
4C2
Total number of matches,
6C2 + 6C2 + 6C2 + 4C2 + 2
= 53

## Quantitative Aptitude for SSC CGL

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests