MCQ: Permutations and Combinations - 1 - SSC CGL MCQ

MCQ: Permutations and Combinations - 1 - SSC CGL MCQ

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15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Permutations and Combinations - 1

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MCQ: Permutations and Combinations - 1 - Question 1

In how many ways 5 girls and 3 boys be seated in a row, so that no two boys are together?

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

Number of Girls = 5
Number of Boys = 3
Total ways in which 5 girls can be seated
5P5 = 5 x 4 x 3 x 2 x 1 = 5! ways
The 5 girls are placed as shown below,

Here, 6 places are possible for boys such that no boys sit together.
So, the ways in which boys can sit.
⇒ 6P3
So, overall number of possible seating arrangements,
5P5 x 6P3 = 5 x 4 x 3 x 2 x 1 x 6 x 5 x 4
5P5 x 6P3 =14400 ways
Thus, total required ways are 14400.
Hence, the correct option is 2.

MCQ: Permutations and Combinations - 1 - Question 2

A volleyball team has 6 people. In how many different ways can teams be made from a class of 12 people?

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

Calculations
We can choose 6 people from a Volleyball team.
The no. of ways of choosing  6 Players from among 12 players is = 12C6

∴ Option (2) is the correct answer.

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MCQ: Permutations and Combinations - 1 - Question 3

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 3

Given:
5 men selected from 8 men
6 women selected from 10 women

Formula used:
nCr = n!/[r!(n - r)!]
Where n = possible outcome
r = required outcome

Calculation:
Ways to select 5 men
8C5 = 8!/[5!(8 - 5)!]
⇒ (6 × 7 × 8)/(3 × 2) = 56
Ways to select 6 women
10C6 = 10!/[6!(10 - 6)!]
⇒ (7 × 8 × 9 × 10)/(4 × 3 × 2) = 210
Ways to select 5 men and 6 women
56 × 210 = 11760
∴ 5 men and 6 women can be selected in 11760 ways.

MCQ: Permutations and Combinations - 1 - Question 4

In how many different ways can the letters of the word 'GEOGRAPHY' be arranged such that the vowels must always come together?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 4

Given:
The given number is 'GEOGRAPHY'

Calculation:
The word 'GEOGRAPHY' has 9 letters. It has the vowels E, O, A in it, and these 3 vowels must always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, GGRPHY(EOA).

Let 7 letters in this word but in these 7 letters, 'G' occurs 2 times, but the rest of the letters are different.

Now,
The number of ways to arrange these letters = 7!/2!
⇒ 7 × 6 × 5 × 4 × 3 = 2520
In the 3 vowels(EOA), all vowels are different
The number of ways to arrange these vowels = 3!
⇒ 3 × 2 × 1 = 6

Now,
The required number of ways = 2520 × 6
⇒ 15120
∴ The required number of ways is 15120.

MCQ: Permutations and Combinations - 1 - Question 5

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that atleast 3 men are there on the committee. In how many ways can it be done ?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 5

Given:
(7 men + 6 women) 5 persons are to be chosen for a committee.

Formula used:
nCr = n!/(n - r)! r!

Calculation:
Ways in which at least 3 men are selected;
⇒ 3 men + 2 women
⇒ 4 men + 1 woman
⇒ 5 men + 0 woman

Number of ways = 7C3 × 6C2 + 7C4 × 6C1 + 7C5 × 6C0
⇒ 7!/(3! × 4!) × 6!/(2! × 4!) + 7!/(4! × 3!) × 6!/(1! × 5!) + 7!/(5! × 2!) × 6!/(6!× 0!)
⇒ 35 × 15 + 35 × 6 + 21
⇒ 735 + 21 = 756
∴ The required no of ways = 756.

MCQ: Permutations and Combinations - 1 - Question 6

In how many ways can the word CHRISTMAS be arranged so that the letters C and M are never adjacent?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 6

Given:
Different words from CHRISTMAS have to be formed.

Formula:
Words in which letters C and M are never adjacent = All cases – words having C and M together.

Calculation:
⇒ Total number of words = 9!/2! (Division of 2! As S is repeated)
Let us assume C and M to be one unit. Then, letters can be arranged in 8! Ways. C and M can be arranged in 2! Ways. Letter S is repeated, so total number of ways will be divided by 2!
⇒ Number of words with C and M adjacent= 8!/2! × 2! = 8!
Words in which letters C and M are never adjacent = 9!/2! – 8! = 8! × (9/2 - 1) = 8! × (7/2)

MCQ: Permutations and Combinations - 1 - Question 7

How many four-digit numbers can be formed with digits 2, 5, 6, 7 and 8? (Repeating digits are not allowed)

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 7

Given:
5 number are given 2, 5, 6, 7 and 8
Four-digit number without repetition
Formula used:
Permutation for no repetition
Where n = total possible numbers
r = required number

Calculation:
Here the total possible number n = 5
And Required number r = 4
Applying the formula

⇒ 5!
⇒ 5 × 4 × 3 × 2 × 1 = 120
∴ There will 120 possible four-digit number.

MCQ: Permutations and Combinations - 1 - Question 8

How many different 6-digit numbers can be formed from the digits 4, 5, 2, 1, 8, 9 ?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 8

The given digits are 4, 5, 2, 1, 8, 9
Therefore required number of ways

Hence, the correct answer is 720.

MCQ: Permutations and Combinations - 1 - Question 9

If 2nC3 : nC2 = 12 : 1, then the value of n is ?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 9

Given:
If 2nC3 : nC2 = 12 : 1

Formula used :
nCr = n!/r!(n - r)!

Calculation:
If 2nC3 : nC2 = 12 : 1    ----(1)
Using the formula
2nC2 = {2n!/3! (2n - 3)!}
⇒ 2n(2n - 1)(2n - 2)(2n - 3)!/(2n - 3)! × 3 × 2
⇒ n(2n - 1)2(n - 1)/3      ----(2)
nC2 = {n!/2! (n - 2)!
⇒ n(n - 1)(n - 2)!/(n - 2)! × 2
⇒ n(n - 1)/2      ----(3)
Putting equation 2 and 3 in equation 1
⇒ {n(2n - 1)2(n - 1)/3}/{n(n - 1)/2} = 12/1
⇒ 2n - 1 = 9
⇒ 2n = 10
⇒ n = 5
∴ The value of n is 5.

MCQ: Permutations and Combinations - 1 - Question 10

How many possible two-digit numbers can be formed by using the digits 3, 5 and 7 (repetition of digits is allowed)?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 10

⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3.
∴ 9 possible two-digit numbers can be formed.
The 9 possible two-digit numbers are:
33, 35, 37, 53, 55, 57, 73, 75, 77

MCQ: Permutations and Combinations - 1 - Question 11

How many 3 digit odd numbers can be formed from the digits 5, 6, 7, 8, 9, if the digits can be repeated

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 11

Given:
5, 6, 7, 8, 9 are the digits to form 3 digit number

Calculation:
Let us take the 3digit number as H T U (Hundreds, tens, unit digit)  respectively
To make 3 digit number as odd
5, 7, 9 are only possibly be used in the unit digit place
In hundreds and tens place all  5 digits are possible
Number of ways for unit digit = 3
Number of ways for tens digit = 5
Number of ways for hundreds digit = 5
Number of 3 digits odd number =  3 × 5 × 5 = 75
∴ 75 Three-digit odd numbers can be formed from the digits 5, 6, 7, 8, 9 if the digits can be repeated

MCQ: Permutations and Combinations - 1 - Question 12

In how many different ways can the letters of the word 'FIGHT' be arranged?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 12

Given
Total alphabets in word 'FIGHT' = 5

Concept Used
Total number ways of arrangement = n!

Calculation
The number of different ways of arrangement of n different words (without repetition) = 5!
⇒ 5 × 4 × 3 × 2 × 1 = 120
∴ The required answer is 120

MCQ: Permutations and Combinations - 1 - Question 13

In how many ways can we sort the letters of the word MANAGEMENT so that the comparative position of vowels and consonants remains the same as in MANAGEMENT.

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 13

Given:
Word = MANAGEMENT

Calculation:
Vowel occupy 4 places then !4
∵ A and E are repeated then !4/(!2 × !2)
Consonant occupy 6 places then !6
⇒ M and N are repeated then !6/(!2 × !2)

= 6 × 180 = 1080

MCQ: Permutations and Combinations - 1 - Question 14

The number of ways of arrangements of 10 persons in four chairs is -

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 14

Given:
The number of ways of arrangements of 10 persons in four chairs

Formula used:
nPr = n!/(n – r)!
Where, n = Number of persons
r = Number of chairs

Calculation:
According to the question
nPr = n!/(n – r)!
⇒ 10!/(10 – 4)!
⇒ 10!/6!
⇒ (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(6 × 5 × 4 × 3 × 2 × 1)
⇒ (10 × 9 × 8 × 7)
⇒ 5040
∴ The required value is 5040

MCQ: Permutations and Combinations - 1 - Question 15

In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?

Detailed Solution for MCQ: Permutations and Combinations - 1 - Question 15

Taking all person of same nationality as one person, then we will have only three people.
These three person can be arranged themselves in 3! Ways.
8 Indians can be arranged themselves in 8! Way.
4 American can be arranged themselves in 4! Ways.
4 Englishman can be arranged themselves in 4! Ways.
Hence, required number of ways = 3! 8! 4! 4! Ways.

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Quantitative Aptitude for SSC CGL

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