Important Questions Class 11 Maths Chapter 7

Important Questions Class 11 Mathematics Chapter 7

Important Questions for CBSE Class 11 Mathematics Chapter 7 – Permutations and Combinations

Permutation refers to the various ways a number or set of things can be arranged, whereas combination refers to a specific arrangement of distinct elements. The permutation is introduced to students in Chapter 7 of Class 11 Mathematics. Since they will be studying the topic for the first time, it is important to have a good grasp of it. The major topics that this chapter covers are:

  • Fundamental Principles of Counting Permutations 
  • Permutations
  • Permutation when all objects are distinct
  • Factorial notation 
  • Derivation of the formula for nPr
  • Permutations when all the items are not separate objects 
  • Combinations

Extramarks’ Important Questions for CBSE Class 11 Mathematics Chapter 7 gives students the questions that are most likely to be asked in exams. These questions cover the main topics of the chapter and are compiled by subject matter experts by referring to NCERT and past years’ question papers. These questions can be useful for students who prefer studying in a question-answer format, which is also easier for revision. 

CBSE Class 11 Mathematics Chapter-7 Important Questions

Study Important Questions for Class 11 Mathematics Chapter-7 – Permutations and Combinations

Given below are Important Questions for Class 11 Mathematics Permutations and Combinations each for 1, 2, 4 and 6 marks.

1 Mark Questions

Q1. Given 5 different coloured flags, how many different signals can be generated if each signal uses two flags one beneath the other?

Ans.

The first flag can be selected in one of five ways. There are four options for selecting the second flag.

According to F.P.C., the total number of ways is 54=20.

Q2. The value of 0! Is?

Ans. 

0!= 1

2 Marks Questions

Q1. How many integers between 100 and 1000 can be produced with the digits 0, 1, 2, 3, 4, 5 if digit repetition is not permitted?

Ans: All numbers between 100 and 1000 are three digits. To begin, we must count six-digit permutations three at a time. The number would be 6P3. These permutations, however, will include those in which 0 is placed in the 100th position. For example, 092, 042, and so on are two-digit numbers from which the number of such numbers must be subtracted. 6P3 to calculate the correct number. We place a 0 at the 100th position and rearrange the remaining numbers to find the total number of such numbers.

=6P3 – 5P2 = 6!3! – 5!3!

=4×5×6-4×5 = 100

Q2. A coin toss’s outcome is recorded six times. How many possible outcomes are there?

Ans. Each individual outcome in the coin toss problems is regarded as a record of the coin flips in the order in which they occurred. Despite the fact that both contain “HHTHTT,” “HTTTHH” is a distinct result. They come in threes. Multiplication is required because each coin flip has two options and we are flipping the coin six times.

According to the rule of thumb, there will be: 22222222=64

4 Marks Questions

Q1. Determine the number of different words that can be formed using only the letters of the word TRIANGLE, with no vowels in the same place.

Ans. First, arrange the consonants.

There are five ways to arrange five consonants.

There are now six spaces between these five consonants in which to arrange vowels. As a result, there are 6P3 ways to arrange the vowels.

Dot positions should be used for consonants. There are a total of 5! = 120 different ways to do this.

The number of cross spots is six.

If we put vowels in these places, no two vowels will be together.

This can be done in nCr = 654 = 120 different ways.

The required number of ways is 120120, which equals 14400.

Q2. Calculate the number of 5 card combinations from a deck of 52 cards if each combination has exactly one ace.

Ans. A 52-card deck contains four aces.

A five-card combination with exactly one ace is required.

Following that, one ace can be chosen.

The remaining four cards in the deck can be chosen in a 4C1 from the deck’s 48 cards.

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

= 48C1×4C1

48×47×46×45 4×3×2×1×4=778320

6 Marks Questions

Q1. How many different ways can ten books be arranged on a shelf so that a particular pair of books is never found together?

Ans. The total number of books is ten.

There are a total of ten methods!

Consider a pair of books to be a single total set =9.

There are approximately nine! different ways to arrange them.

It is possible to arrange two books in two!

The total number of pairings for the two books is 9!2!

The number of ways that are not together equals 10!-9!2! = 9!10 – 2 = 8 9! ways

Q.1 If  nC0+ nC1+ nC2+ ….. nCn = 128, then 8Cn is equal to

16.

8.

7.

1.

Marks:1

Ans

 nC0+ nC1+ nC2+ ….. nCn = 1282n=27=7Therefore, 8Cn+ 8C7= 8C1=8

Q.2 If a polygon has 20 diagonals, the number of its sides is

8.

9.

10.

12.

Marks:1

Ans

8.

The number of diagonals for n sided polygon = nC2n=n(n3)2We have, n(n3)2=20n(n3)=40n23n40=0n28n+5n40=0n(n8)+5(n8)=0(n8)(n+5)=0n=8orn=5(neglected)So number of diagonals = 8

Q.3 Every body in a room shakes hands with everybody else. The total number of handshakes is 66. Find the number of persons in the room.

Marks:4

Ans

Let the number of persons in the room be n Since one person at a time can make hand shake with 1 person only. The total number of hand shakes=The total number of selection of any 2 person out of n person=nC2µnC2=66 (given)+n2(n+2)=66+n(n+1)(n+2)2(n+2)=66+n(n+1)2=66+n(n+1)=2×66+n(n+1)=2x11x6+n(n+1)=12x11On comparing factors, we get n=12The number of persons in the room is 12.

Q.4 Find the value of n if  (i) 2nC3 : nC3 = 12 : 1 (ii) 2nC3 : nC3 = 11:1

Marks:4

Ans

(i) 2nC3:nC3=12:12nC3nC3=12:1+2n3(2n+3)n3(n+3)=1212n(2n+3)x(n+3)n=1212n(2n+1)(2n+2)(2n+3)(2n+3)x(n+3)n(n+1)(n+2)(n+3) =121+2n+1=3(n+2)+n=5

(ii) 2nC3:nC3=11:1C32nC3n=111+2n3(2n+3)n3(n+3)=111+2n(2n+3)x(n+3)n=111+2n(2n+1)(2n+2)(2n+3)(2n+3)x(n+3)n(n+1)(n+2)(n+3) =111+4(2n+1)(n+2)=111+n=6

Q.5 Find the number of different arrangements for letters of the word EQUATION. In how many ways, do the vowels come together?

Marks:4

Ans

There are 8 different letters in the word EQUATION
This can be arranged in 8 = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ways
= 40320 ways
There are 5 vowels and we will assume it as a single letter.
Now, bundle of 5 vowels and letters Q, T, N that is 4 different letters can be arranged in 4 = 4 x 3 x 2 x 1 = 24 ways.
The 5 different vowels can be arranged among itself in 5 = 5 x 4 x 3 x 2 x 1 = 120 ways.
Thus, the total number of arrangement in which vowels come together = 120 x 24 = 2880 ways.

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FAQs (Frequently Asked Questions)

Permutations are defined as the orderly arrangement of objects or numbers. Combinations are defined as the selection of objects or numbers from a group in such a way that the order is not disturbed. This is an important chapter that should be practised thoroughly with clear concepts. This is easily understood with the proper practise of the various numerical systems.

Given that repetition is not permitted, any of the five numbers can take the place of the unit. As a result, the number of possible unit placements is 5. The tens can then be taken by resting four numbers and the hundreds by resting three digits on the left. As a result of the multiplication principle, the final result is 5 x 4 x 3 = 60.

The following combinations can be found with the first 10 letters of the English alphabet to form a four-letter code:

  • The first letter can be used in ten different ways.
  • The second letter can be used in nine different ways.
  • The third letter can be used in eight different ways.
  • The fourth letter can be used in seven different ways.
  • Thus, according to the fundamental principle of counting, 10.9.8.7=5040.

 

Students can pick up your numericals and practise this way. This allows you to understand the topic’s fundamental principles. To be implemented, the formulas must be learned and understood.