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

Given:

We are asked to find how many 3-digit even numbers can be formed with no digit repeated using the digits 0, 1, 2, 3, 4, and 5.

Solution:

For a 3-digit number to be even, its unit (last) digit must be one of the even digits. In this case, the even digits available are 0, 2, and 4.

Case 1: The unit digit is 0

If the unit digit is 0, then the first digit (hundreds place) can be chosen from the remaining digits 1, 2, 3, 4, and 5. There are 5 choices for the first digit.

The second digit (tens place) can be chosen from the remaining 4 digits, giving us 4 choices for the second digit.

Thus, the total number of even 3-digit numbers when the unit digit is 0 is:

5 × 4 = 20

Case 2: The unit digit is 2

If the unit digit is 2, then the first digit (hundreds place) can be chosen from the remaining digits 1, 3, 4, and 5. There are 4 choices for the first digit.

The second digit (tens place) can be chosen from the remaining 4 digits (0, 1, 3, 4, 5), giving us 4 choices for the second digit.

Thus, the total number of even 3-digit numbers when the unit digit is 2 is:

4 × 4 = 16

Case 3: The unit digit is 4

If the unit digit is 4, then the first digit (hundreds place) can be chosen from the remaining digits 1, 2, 3, and 5. There are 4 choices for the first digit.

The second digit (tens place) can be chosen from the remaining 4 digits (0, 1, 2, 3, 5), giving us 4 choices for the second digit.

Thus, the total number of even 3-digit numbers when the unit digit is 4 is:

4 × 4 = 16

Total number of 3-digit even numbers:

20 (unit digit 0) + 16 (unit digit 2) + 16 (unit digit 4) = 52

Answer: 52

Let be the number of girl players.
Let be the number of boy players.

. There are 14 girl players.

. There are 16 boy players.

Number of matches between each boy and each girl:

Therefore, the required answer is 224.

Explanation:

The word 'SRIRAJASTHANAMA' has 9 different letters

S R I A J T H M T.

The number of different 5 letter' words, with or without meaning

= 9C5 × 5!

Option (3) is true.

Given:

The word is COVID

Number of letters = 5

Formula used:

Number of different words = n!

Where n = number of letters

Calculation:

Number of different words = 5!

⇒ Number of different words = 5 × 4 × 3 × 2 × 1

⇒ Number of different words = 120

∴ The correct answer is option (4).

Explanation:

Given,

Number of courses, n > 4

The student needs to choose (n - 2) courses out of n courses, with 2 courses being compulsory

∴Total courses is n – 2 and total remaining courses that can be chosen from (n – 2) courses is ( n – 4).

∴ Required number of ways ^(n-2) C _(n-4).

⇒ 
 

∴ Option d is correct.
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⇒ 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.