MCQ: Permutations and Combinations - 1 - SSC CGL MCQ

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.

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.

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

Formula used:
ⁿC_r = 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.

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.

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

Formula used:
ⁿC_r = 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.

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)

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.

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

Hence, the correct answer is 720.

Given:
If ²ⁿC₃ : ⁿC₂ = 12 : 1

Formula used :
ⁿC_r = 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.

⇒ 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 

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

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 

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

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

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

Calculation:
According to the question
ⁿP_r = 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

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.