Permutation and Combination MCQ Quiz - Objective Question with Answer for Permutation and Combination - Download Free PDF

Given:

Number of distinct points = 15

Formula used:

Maximum number of triangles = C(n, 3)

Where, C(n, r) = 

Calculation:

Maximum number of triangles = C(15, 3)

⇒ C(15, 3) = 

⇒ C(15, 3) = 

⇒ C(15, 3) = 455

∴ The correct answer is option (2).

Given:

A 6-digit security code is made using digits from 0 to 9. The first and the last digits are known. The remaining four digits are prime numbers.

Prime numbers between 0 and 9: 2, 3, 5, 7

Formula used:

Number of trials = Total possible combinations of the four prime digits

Calculation:

The total number of choices for each of the four remaining digits = 4 (since there are 4 prime numbers: 2, 3, 5, 7)

Total combinations = 4 × 4 × 4 × 4

⇒ Total combinations = 4⁴

⇒ Total combinations = 256

∴ The correct answer is option (3).

Given:

Digits available: 1, 2, 3, 4, 5, 6

Requirement: Form 4-digit numbers

Calculations:

Let the 4-digit number be represented by four positions:

Since the string 123 must appear and no digit can be repeated, the digits 1, 2, and 3 are used up when the string "123" is placed.

The remaining available digits are {4, 5, 6}.

There are two possible ways for the string 123 to appear in a 4-digit number:

Case 1: 123 occupies the first three positions.

The number format is 1 2 3 _

The first three digits are fixed as 1, 2, and 3.

The fourth position (the blank) must be filled by one of the remaining digits from {4, 5, 6}.

Number of choices for the fourth digit = 3 (either 4, 5, or 6).

The possible numbers are: 1234, 1235, 1236.

Number of numbers in Case 1 = 3.

Case 2: 123 occupies the last three positions.

The number format is _ 1 2 3

The last three digits are fixed as 1, 2, and 3.

The first position (the blank) must be filled by one of the remaining digits from {4, 5, 6}.

Number of choices for the first digit = 3 (either 4, 5, or 6).

The possible numbers are: 4123, 5123, 6123.

Number of numbers in Case 2 = 3.

Total number of 4-digit numbers = (Numbers from Case 1) + (Numbers from Case 2)

⇒ Total number = 3 + 3 = 6

∴ The correct answer is option 3.

Given:

Digits: 1, 2, 3, 4, 5, 6, 7

Total digits in each number: 4

'123' must always appear as a string.

Formula used:

Number of arrangements = (Number of positions for the fixed string) × (Number of ways to arrange the remaining digits)

Calculation:

'123' is treated as a single unit (fixed string). Remaining digits: 4, 5, 6, 7 (4 digits).

Possible positions for the fixed string '123' in a 4-digit number:

Positions: [ _ _ _ _ ]

'123' can occupy positions: 1-3 or 2-4 (2 positions).

For each position of '123', the remaining digit can occupy the 4th position.

Number of arrangements of the remaining digits = 4! / (4-1)! = 4.

⇒ Total numbers = 2 × 4 = 8

∴ The correct answer is option 1.

Given:

Word: WORKSPACE

Total letters = 9

Vowels = O, A, E (3 vowels)

Consonants = W, R, K, S, P, C (6 consonants)

Condition: Vowels must always be together

Calculations:

If vowels are together, treat them as one unit

So total units = 6 consonants + 1 vowel group = 7 units

Ways to arrange these 7 units = 7!

Ways to arrange 3 vowels within the group = 3!

7! = 5040

3! = 6

Total ways = 5040 × 6 = 30240

∴ Total arrangements = 30240

⇒ 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:

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.

Explanation:

Total people in meeting is 45 and out of then 40 people know each other.

So 5 people don't know anyone.

Let those 5 peoples be A, B, C, D, E

So A will handshake 44 people.

B will handshake 43 people

C will handshake 42 people

D will handshake 41 people

and E will handshake 40 people

Hence total handshake = 44 + 43 + 42 + 41 + 40 = 210

Option (3) is correct

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.

Important PointsThe value of 0! is 1.

448 = 2 × 2 × 2 × 2 × 2 × 2 × 7

⇒ 448 = 2⁶ × 7¹

∴ The required number of mobile phones be shared equally in the students = (6 + 1) × (1 + 1) = 7 × 2 = 14

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:

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:

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:

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.