Important Questions For Class 11 Maths Chapter 7: Permutations and Combinations

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Important Questions for Class 11 Maths Chapter 7 Permutations and Combinations are provided in the article. Permutation and combination are two different approaches to representing a set of items. Combination is when the order of the elements doesn't matter. A permutation is often referred to as an ordered combination. It describes various ways in which data can be organized by forming subsets. Basic counting methods include permutation and combinations. The figures at the end of each potential data arrangement and combination using P & C.

Also read: Isosceles Triangle Theorems

Very Short Answer Type Questions (1 Mark Questions)

Ques. Evaluate the value of: 4! – 2!

Ans. 4! = 1×2×3×4=24

2! = 1 × 2 = 2

Therefore= 4! – 2! = 24 – 2= 22

Ques. Determine the three-digit numbers that may be formed from the given numbers: 1, 2, 3, 4, and 5, assuming that the numbers can be repeated.

Ans. There are now five options for filling one's place. Because repetition is possible, the tens spot can be filled in five different ways. Similarly, hundreds of places can be filled in five different ways. As a result, the number of possibilities to make three-digit numbers from the provided digits is 5×5×5=125.

Ques. If letter repetition is not permitted, how many 3-letter words even without meaning can be created from the letters in the word 'LOGARITHMS'?

Ans. There are ten letters in the word LOGARITHMS.

As a result, the number of three-letter words with or without meaning created by combining these letters has increased.

The first letter can be chosen in 10 different ways.

There are 9 different ways to choose the second letter

Three letters can be chosen in 8 different ways

The total amount of words is 10 × 9 ×8 = 720 words

Ques. There are 10 chair patterns and eight table layouts available to an event planner. How many different ways can he create a set of tables and chairs?

Ans. The event planner has 10 chair designs and 8 table styles.

There are 10 different ways to choose a chair.

There are 8 different ways to choose a table.

As a result, one chair and one table may be chosen in 10×8 = 80 different ways.

Ques. Make factorials out of the following products: 4×5×6×7×8×9×10.

Ans. 1×2×3×4×5×6×7×8×9×10×2×3

= 10!3!

Ques. How many rectangles may be constructed on a chessboard if there are 9 horizontal lines and 9 vertical lines?

Ans. Total number of rectangle = 9C2 × 9C2

= 36 × 36 = 1296

Also read: Area of a Triangle

Short Answer Type Questions (2 Marks Questions)

Ques. If digit repetition is not permitted, how many integers between 100 and 1000 may be produced with the digits 0, 1, 2, 3, 4, 5?

Ans. All numbers in the range of 100 to 1000 are three-digit numbers. To begin, we must count the permutations of six digits 3 at a time. 6P3 would be the number. However, these permutations will include ones in which 0 is placed in the 100th position. For example, 092, 042, and so on are 2-digit numbers, and the number of such numbers must be deducted from. 6P3 to arrive at the appropriate number. We fix 0 at the 100th spot and rearrange the remaining numbers to determine the total number of such numbers.

=6P3 - 5P2 = 6!3! - 5!3!

=4×5×6-4×5 = 100

Ques. You need to assemble a committee of 5 gentlemen and 6 women but there are 8 men and 10 women. Find out different ways in which a committee can be formed?

Ans. We must choose 5 gentlemen from a group of 8 gentlemen and 6 women from a group of 10 women. There are several ways for it, much like,

= 8 C5×10 C6

= 8 C5 × 10 C6 nCr = nCn-r

= [(8 x 7 x 6)/(3 x 2 x 1)] x [(10 x 9 x 8 x 7)/(4 x 3 x 2 x 1)]

= 56 × 210

= 11760

Also read: Involute

Ques. The outcomes of a coin toss are recorded 6 times. How many different outcomes are there?

Ans. We regard each individual outcome in the coin toss problems as a record of the coin flips in the order they occurred. Despite the fact that both included "HHTHTT”, "HTTTHH" is a unique result. There are three of them. Because each coin flip has two choices and we are flipping the coin six times, multiplication is required.

According to principle, there will be: 2×2×2×2×2×2=64

Ques. How many different ways can a team of 5 people be established from a total of 10 people so that two specific people are included in each team?

Ans. Each team should have two particular individuals on it. As a result, we must choose the remaining (5 – 2) = 3 people from a total of (10 – 2) = 8 people. As a result, the necessary number of ways has been met.

= 8C3

= (8×7×6)(3×2×1)

= 8×7

= 56

Ques. When 9 courses are offered and two specific courses are mandatory for every student, how many ways may a student pick 5 programmes?

Ans. There are nine courses available, with two of them being mandatory for all students. As a result, each student must select three courses from the remaining seven. This can be selected in 7C3 ways. As a result, the maximum number of programme selection options is met.

7C3=7!3!4!

=7 × 6 × 53 × 2 × 1

= 35

Ques. What is the maximum number of chords that can be drawn across 21 points on a circle?

Ans. Any two points on a circle can be joined to form a chord. As a result, the total number of chords drawn via 21 points equals the number of ways to choose two points from 21 points. This can be accomplished in 21C2.

As a result, the total number of chords= 21C2 = 21!19!2! = 21×10= 210

Also Check:

Chapter Related Topics
Parabola Formula	Permutations and combinations formulas	Factorial Formula
Stirling formula	Difference between permutation and combination	Combinatorics

Long Answer Type Questions (3 Marks Questions)

Ques. If each combination has precisely one ace, calculate the number of 5 card combinations from a deck of 52 cards.

Ans. There are four aces in a deck of 52 cards.

It's necessary to make a five-card combination with exactly one ace.

After that, one ace can be chosen.

Out of the 48 cards in the deck, 48C4 the remaining four cards can be chosen in a 4C1

As a result, the needed number of 5 card combinations may be calculated using the multiplication principle.

= 48C1×4C1

48×47×46×45 4×3×2×1×4=778320

Ques. The number of ways that 2 black and 3 red balls can be picked from a bag containing 5 black and 6 red balls is determined by the number of black and red balls in the bag. find the total no of choices.

Ans. The number of black balls is equal to 5.

The number of red balls is 6.

The number of black balls to be chosen is 2.

The number of red balls to be chosen is 3.

Total number of choices = 5C2 × 6C3

5!5 - 2!2 × 6!6 - 3!3

5 × 4 × 3!3! × 2 × 6 × 5 × 4 × 3!3! × 3 × 2 = 200

Ques. Find the number of alternative words that can be made using only the letters of the word TRIANGLE, with no vowels in the same place.

Ans. Firstly arrange the consonant

5 consonants can be arranged in 5 ways.

Now we have six spaces between these five consonants to arrange vowels. Therefore, the number of ways to arrange the vowels are 6P3.

Consonants should be placed in dot positions. This may be done in a total of 5! = 120 different ways.

Number of cross spots = 6

If we put vowels in these spots, then no 2 vowels are together.

This can be done in nCr ways =6×5×4 = 120 ways

Required number of ways =120×120 = 14400

Ques. A little child has 3 library tickets and 8 books in the library that he is interested in. He does not wish to borrow Chemistry part II unless Chemistry part I is also acquired out of these 8. He has a variety of options for selecting the 3 books to be borrowed.

Ans. The following are the many ways to borrow three books:

Considering that chemistry part II is selected and Chemistry part I is also borrowed, then the third book is to be selected from the six volumes that are left.
In the event that Chemistry Part II is not picked, he must choose three books from the other seven.

First choice 6C1 = 6 ways.

Second choice 7C3 = 7×6×51×2×3 = 35 ways.

The total number of ways he may pick the three books he wants to borrow is = 6 + 35 = 41

Ques. Everyone at a party shakes hands with everyone else. Determine the number of people that attended the party if there were a total of 105 handshakes.

Ans. Let no. of persons be 'n' then

No. of handshakes = nC2

= nC2=105

=n!(n-2!)=210=n2-n-210 =0

=n=15 (n>0)

Also read: Difference between Sequence and Series

Very Long Answer Type Questions (5 Marks Questions)

Ques. If n = 20 and r = 4, calculate the number of permutations and combinations.

Ans. Permutation Formula = nPr

nPr = n!n-r!

= 20p4

=20!/(20-4)!

= 20!16!

= 20×19×18×17×16!16!

= 116280

Combination Formula = nCr

nCr = n!(n-r)!r!

= 20C4

= 20!(20-4)!4!

= 20!16!4!

= 20×19×18×17×16!16!4!

= 20×19×18×174×3×2

= 5×19×3×17

= 4845

Ques. There are five vowels and twenty-one consonants in the English alphabet. How many words can be made from the alphabet using two distinct vowels and two different consonants?

Ans.

	Total number	Number to be chosen	Number of ways to choose
Consonant	5	2	5C2
Vowels	21	2	21C2

Number ways of selecting 2-vowels and 2-consonant

= 5C2 × 5C2

= 5!2(5!-2!)×21!2!(21!-2!)

= 5!2!3!×21!2!19!

= 5×4×3!2×1×3!×21×20×19!2×1×19!

= 10 × 210 = 2100

Henceforth, the possible arrangements of the aforementioned 4 letters are,

4P4 = 4!(4-4)! = 4×3×2×1 = 24

Total ways to choose the 4 letters = 24 ways

The total number of words that can be formed is = number of ways to choose two consonants × number of ways to choose two vowels × number of ways to arrange four letters.

∴Total number of words of 4 letters = 5C2 × 21C2 × 4! = 210 × 10 × 24 = 50,400 words

Also check: NCERT Solutions for Class 11 Maths

Ques. How many different ways may 10 books be put on a shelf so that a specific pair of volumes are never found together?

Ans. The number of books is 10.

The total number of methods is ten!

Given a pair of books, treat them as a single total set =9.

There are about 9! different ways to arrange them.

Two books can be arranged in 2!

The total number of ways that the two books are paired =9!×2!

Number of ways that are not together= 10!-9!×2! = 9!10 - 2 = 8 × 9! ways

Also Read:

Important Math Topics
Math MCQs	Mensuration	Trigonometry
Number Systems	Probability and Statistics	Arithmetic
Algebra	Difference between topics in maths	Math Study Material

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