 # RBSE Maths Class 11 Chapter 7: Important Questions and Solutions

RBSE Maths Chapter 7 – Permutation and combination Class 11 Important questions and solutions are available here. The important questions and solutions of Chapter 7, available at BYJU’S, contain step by step explanations. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 11 solutions.

Chapter 7 of the RBSE Class 11 Maths will help the students to solve problems related to the fundamental principles of multiplication and addition, permutations, number of permutations, circular permutations, clockwise and anticlockwise permutations, combinations. After referring to these important questions and solutions, you can refer to 2020-2021 solutions for the chapter here.

### RBSE Maths Chapter 7: Exercise 7.1 Textbook Important Questions and Solutions

Question 1: Find the value of n.

[i] n-1P3 : n+1P3 = 5:12

[ii] nP6 = 10 * nP5

[iii] 56Pr+6 : 54Pr+3 = 30800

[iv] 6+nP2 : 6-nP2 = 56:12

Solution:

[i] n-1P3 : n+1P3 = 5:12 The value n = 8 is taken as n must be a whole number and not a fraction.

Hence n = 8.

[ii] nP6 = 10 * nP5

nP6 = 10 * nP5

n! / (n – 6)! = 10 * [n! / (n – 5)!] [1 / (n – 6)!] = 10 * [1 / (n – 5)!] [1 / (n – 6)!] = 10 * [1 / (n – 6)! * (n – 5)]

1 = 10 * (1 / (n – 5))

n – 5 = 10

n = 10 + 5 = 15

[iii] 56Pr+6 : 54Pr+3 = 30800

56Pr+6 = 56! / [56 – (r + 6)!] = 56! / (50 – r)!

54Pr+3 = 54! / [54 – (r + 3)!] = 54! / (51 – r)!

56Pr+6 / 54Pr+3

= [56! / (50 – r)! * 54! / (51 – r)!] (56 * 55 * 54! / (50 – r)!) * ((51 – r) * (50 – r)! / 54!)

= 56 * 55 * (51 – r)

56 * 55 * (51 – r) = 30800

51 – r = 10

r = 41

[iv] 6+nP2 : 6-nP2 = 56:12

P (6 + n, 2) : P (6 – n, 2) = 56:12 = 14:3

[(6 + n) (6 + n – 1)] : [(6 – n) (6 – n – 1)] = 14:3

[(6 + n) (5 + n)] : [(6 – n) (5 – n)] = 14:3

(30 + 11n + n²) : (30 – 11n + n²) = 14:3

90 + 33n + 3n² = 420 – 154n + 14n²

11n² – 187n + 330 = 0

n² -17n + 30 = 0

(n -15) (n – 2) = 0

n = 2, 15

Question 2: Find the number of different words formed by letters of the word ALLAHABAD.

Solution:

There are 9 letters in the given word, in which 4 As, 2 Ls are present and others are different. So, by taking all letters, number of different words that can be formed are

= [9!] / [4! 2!]

= [9 * 8 * 7 * 6 * 5] / 

= 15120 / 2

= 7560.

Question 3: How many words can be formed using letters of the word TRIANGLE? Out of these how many words begin with T and end with E?

Solution:

There are 8 letters in the given word, in which there is no repetition of any letter.

Hence, the number of words formed by these letters are = (8)! = 40320

If we fix the position of T in the starting and E at the end, then the number of remaining letters = 6.

Hence, the number of words formed by these letters = (6)! = 720

Question 4: How many numbers lying between 3000 and 4000 can be formed with the digits 1, 2, 3, 4, 5, 6 which are divisible by 5?

Solution:

The numbers lying between 3000 and 4000 which are divisible by 5 and have 3 on hundredth’s place and 5 on one’s place and the total number of digits is 4.

Total number of numbers = 4P2

= 4! / [4 – 2]!

= 4! / 2!

= [4 * 3 * 2 * 1] / [2 * 1]

= 4 * 3

= 12

12 numbers can be formed.

Question 5: How many 6-digit numbers can be formed by using the digits 0, 1, 2, 3, 4, 5?

Solution:

Given digits = 0, 1, 2, 3, 4, 5.

Hence, 6-digit numbers formed by these digits = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

If zero comes in first, then the numbers become a 5-digit number.

The 5-digit number formed is = 5! = 5 × 4 × 3 × 2 × 1 = 120

Hence, the numbers that do not have zero, in the beginning, are = 720 – 120 = 600.

Question 6: How many 3-digit numbers less than 1000 can be formed using the digits 1, 2, 3, 4, 5, 6 if no digits are repeated?

Solution:

Number of digits given = 6

For forming a 3-digit number, the three-digits from these 6-digits are chosen.

Hence, the total numbers which have 3 digits

= 6P3

= 6! / (6 – 3)!

= [6 * 5 * 4 * 3!] / [3!]

= 6 * 5 * 4

= 120

Question 7: How many ways can 15 members of a committee sit around the round table whereas the secretary sits on one side and vice secretary on the other side of the Director?

Solution:

The total number of members = 15.

Here, the position of secretary, vice secretary and the director is fixed but there is no condition of clockwise and anti-clockwise.

Hence, the total number of sitting ways according to condition = 2!.

Now, change the position of 12 members only.

Hence, the total number of sitting ways of 12 members = 12!.

The total number of ways 15 members sit according to given condition = 12! × 2! = 958003200.

Question 8: There are 15 stations on a railway line. For this, how many different tickets of a class should be printed such that any person from any station can buy the ticket of another station of this line?

Solution:

The total number of stations = 15

A person buys a ticket on a station then we can buy tickets of the remaining 14 stations.

Hence, the total number of methods of printing 14 tickets out of 15 tickets.

= 15P14

= 15! / 14!

= 15 * 14! / 14!

= 15

One ticket can be printed in 2 ways; in this way, a total of 14 tickets will be printed.

Hence, the total number of ways tickets can be printed for 15 stations are = 15 × 14 = 210.

Question 9: In how many ways can 10 different beads be threaded to make a garland such that the 4 special beads never remain separate?

Solution:

4 out of 10 beads do not remain together then, total remaining beads = 10 – 4 = 6

Hence, the methods of making a garland using 6 beads = 6!

If 4 out of 10 beads do not remain together then the method of making a garland = 6! × 4!

As in a garland, there is no difference between clockwise and anticlockwise, so, methods to make a garland with 10 different beads,

= [6! * 4!] / 

= [6 * 5 * 4 * 3 * 2 * 1 * 4 * 3 * 2 * 1] / 2

= 17280 / 2

= 8640

Question 10: How many numbers can be formed using digits 0, 1, 2,…, 9 which is more than or equal to 6000 and less than 7000 and is divisible by 5 whereas any number can be repeated as many times?

Solution:

The numbers which are more than or equal to 6000 and less than 7000 and are divisible by 5 have 6 on thousand’s place and 0 or 5 on one’s place and the number is 4 digits. As digits can be repeated.

 1 10 10 2 5 0 – 9 0 – 9 0,5

Total numbers = 1 × 10 × 10 × 2 = 200.

Question 11: How many words can be formed from letters of the word SCHOOL, whereas both Os do not come together?

Solution:

There are a total of 6 letters in the word SCHOOL.

In it, there are 2 Os and other letters are different, then the number of total permutations = [6! / 2!].

If 2 Os are taken together, then it will be taken as one letter, then the number of permutations taking 2 Os together = 5!.

Hence, the number of permutations when 2 O’s are not together are

= [6! / 2!] – (5!)

= {[6 * 5 * 4 * 3 * 2!] / [2!]} – (5!)

= [6 * 5 * 4 * 3 – 5 * 4 * 3 * 2 * 1]

= 360 – 120

= 240

### RBSE Maths Chapter 7: Exercise 7.2 Textbook Important Questions and Solutions

Question 1: Find the values of n.

[i] 2nC3 : nC3 = 11 : 1

[ii] 20Cn-2 = 20Cn+2

[iii] nC10 = nC15

Solution:

[i] 2nC3 : nC3 = 11 : 1

[2n (2n – 1) (2n – 2)] / [3!] = 11 * [n (n – 1) (n – 2)] / [3!]

4 (2n – 1) = 11 (n – 2)

8n – 4 = 11n – 22

11n – 8n = 22 – 4

3n = 18

n = 6

[ii] 20Cn-2 = 20Cn+2 [iii] nC10 = nC15

nCr = nCn-r

nC10 = nCn-10 = nC15

n – 10 = 15

n = 10 + 15

n = 25

Question 2: Find the value of 50C11 + 50C12 + 51C1352C13.

Solution:

nCr + nCr-1 = (n + 1)Cr —– (1)

50C11 + 50C12 + 51C1352C13

= 51C12 + 51C1352C13

= 52C1352C13

= 0

Question 3: In a triangle ABC, there are 3, 4 and 5 points on side AB, BC, CA respectively. How many triangles will be formed by these points?

Solution:

Using 1 point from each side of the triangle –

3C1 * 4C1 * 5C1 = 3 * 4 * 5 = 60

Using 1 point on 1 side and 2 points on the other side –

(3C1 * 4C2) + (3C1 * 5C2) + (4C1 * 3C2) + (4C1 * 5C2) + (5C1 * 4C2) + (5C1 * 3C2) = 145

On adding = 60 + 145 = 205.

Question 4: A box contains two white, three black and four red balls. Determine the number of ways in which three balls can be selected from this box, in which at least one black ball is compulsory.

Solution:

From the 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that at least one black ball should be there. So, three choices are available.

Total number of ways

= 3C3 + (3C2 * 6C1) + (3C1 * 6C2)

= 1 + [3 * 6] + [3 * (6 * 52 * 1)]

= 1 + 18 + 45

= 64

Question 5: By 6 different coloured flags, taking one or more than one, how many ways can the signal be given?

Solution:

Number of signals using one flag = 6P1 = 6

Number of signals using two flags = 6P2 = 30

Number of signals using three flags = 6P3 = 120

Number of signals using four flags = 6P4 = 360

Number of signals using five flags = 6P5 = 720

Number of signals using all six flags = 6P6 = 720

Therefore, the total number of signals using one or more flags at a time is

= 6 + 30 + 120 + 360 + 720 + 720

= 1956

Question 6: If the number of diagonals in a polygon is 44, then find the number of sides.

Solution:

The number of diagonals of a polygon is given by [n * (n – 3)] / .

Here

[n * (n – 3)] / 2 = 44

n2 – 3n – 88 = 0

(n -11) (n + 8) = 0

n = 11.

The polygon has 11 sides.

Question 7: How many numbers can be formed taking 4 digits from digits 1, 2, 3, 4, 5, 6 whereas digit 4 and 5 are compulsory?

Solution:

Total number of digits = 6

4 digits have to be taken but digits 4 and 5 are compulsory.

Only 2 digits can be chosen out of 6 – 2 = 4 digits.

Now, the numbers formed by using 4 digits is 4P4.

Hence required number

= 4C2 * 4P4

= 6 * 24

= 144

Question 8: In how many ways 6 ‘+’ and 4 ‘-‘ sign can be kept in a line whereas any two ‘-‘ occur together?

Solution:

First, put ‘+’ in a line. This can be done in 1 way because all ‘+’ are the same.

+ + + + + +

Now put ‘-‘ in the following way

– + – + – + – + – + – + –

Hence, the method of putting ‘-‘ sign in 4 out of 7 places = 7

7C1 × 1 because ‘-‘ signs are the same.

Total number of ways 6 ‘+’ and 4 ‘-‘ sign occur together

= 1 * 7C4 * 1

= [7! / 4! 3!]

= 35

Question 9: By 8 students and 5 professors, we have to make a college council, taking 5 students and 2 professors. How many councils can be formed?

Solution:

The number of methods of choosing 5 out of 8 students = 8C5 and number of methods of choosing 2 out at 5 professors = 5C2.

By the basic principle of calculation, the methods of choosing 5 out of 8 students and 2 out of 5 professors are

= 8C5 * 5C2

= [8!] / [5! 3!] * [5! / 2! 3!]

= 8 * 7 * 10

= 560

Hence, required councils are 560.

Question 10: In how many ways can one select a cricket team of 11 from 14 players in which at least 2 bowlers are compulsory whereas only 4 players can bowl. How many ways can this team be formed?

Solution:

The method of choosing 2 out of 4 bowlers is 4C2 and the method of choosing 9 out of the remaining players is 10C9. Similarly, we can find methods for 3 and 4 bowlers.

Hence, the total number of ways to select 11 players out of 14 players

= 4C2 * 10C9 + 4C3 * 10C8 + 4C4 * 10C7

= 6 * 10 + 4 * 45 + 1 * 120

= 60 + 180 + 120

= 360

### RBSE Maths Chapter 7: Miscellaneous Exercise Questions and Solutions

Question 1: If nPn-2 = 60, then n will be:

(A) 2

(B) 4

(C) 5

(D) 3

Solution:

nPn-2 = 60

n! / {n – (n – 2)}! = 60

n! / 2! = 6 * 10

n! = 6 * 2 * 10

n! = 120

n! = 5!

n = 5

Question 2: nPr ÷ nCr is equal to:

(A) n!

(B) (n – r)!

(C) 1 / r!

(D) r!

Solution:

nPr ÷ nCr = [n! / (n – r)!] ÷ [n! / r! (n – r)!]

= [n! / (n – r)!] * [r! (n – r)! / n!]

= r!

Question 3: How many ways 5 persons can sit around the round table:

(A) 120

(B) 24

(C) 60

(D) 12

Solution:

Total number of ways to sitting 5 persons around the round table

= (5 – 1)! = 4!

= 4 * 3 * 2 * 1

= 24.

Question 4: How many words can be formed using letters BHILWARA:

(A) 8! / 2!

(B) 8!

(C) 7!

(D) 6! / 2!

Solution:

There are 8 letters in the given word in which 2 A’s and other letters are different.

Then numbers of words formed by letters of BHILWARA = 8! / 2!.

Question 5: 47C4 +

$$\begin{array}{l}\sum_{r=1}^{5}\end{array}$$
52-jC3 is equal to:

(A) 51C4

(B) 52C4

(C) 53C4

(D) 54C4

Solution: = 52C4

Question 6: Find the value of 61C5760C56.

(A) 61C58

(B) 60C57

(C) 60C58

(D) 60C56

Solution:

61C5760C56

= [61!] / [57! (61 – 57 )!] – [60!] / [56! (60 – 56)!]

= 61! / [57! 4!] – [60!] / [56! 4!]

= 521855 – 487635

= 34220

= 60C57

Question 7: If 15C3r = 15Cr+3, then r is equal to:

(A) 5

(B) 4

(C) 3

(D) 2

Solution:

15C3r = 15Cr+3

nCx = nCy implies that either

i] x = y or

ii] x + y = n

This is because, nCx = [n!] / [x! * (n – x)!] and nCy = [n!] / [y! * (n – y)!]

They will be equal when x = y or n – x = y

n = x + y

Hence,

Case 1 :

3r = 3 + r

=> 2r = 3

=> r = 3 / 2 (fractional answer)

Case 2 :

3r + 3 +r = 15

=> 4r = 15 – 3 = 12

=> r = 3

Question 8: There are 6 points on the circumference of a circle, the number of straight lines joining their points will be:

(A) 30

(B) 15

(C) 12

(D) 20

Solution:

Number of lines passing through n points = nC2

Number of lines passing through 6 points

= 6C2

= 6! / 2! (6 – 2)!

= [6 * 5 * 4!] / [2 * 1 * 4!]

= [6 * 5] / 

= 15

Question 9: How many words can be formed using letters of BHOPAL?

(A) 124

(B) 240

(C) 360

(D) 720

Solution:

Here, the number of letters is 6 and every time we take 6 letters.

Hence, the required number of = 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Question 10: There are 4 points on the circumference of a circle. By joining them how many triangles can be formed?

(A) 4

(B) 6

(C) 8

(D) 12

Solution:

There are three vertices in a triangle.

There are 4 points on the circumference of a circle.

Then, the number of required triangles

= 4C3

= 4! / [3! (4 – 3)!]

= [4 * 3!] / 3! * 1

= 4

Question 11: If nC9 = nC7, then find nC16.

Solution:

nC9 = nC7

nC9 = nCn – 2

On comparing,

9 = n – 7

9 + 7 = n

16 = n

nC16 = 16C16 = 1

Question 12: Find the value of n.

(i) 2nC2 : nC2 = 12 : 1

(ii) 2nC3 : nC3 = 11 : 1

Solution:

(i) 2nC2 : nC2 = 12 : 1

[2n! * (n − 2)! * 2!] / [n! * (2n − 2)! * 2!] = [12 / 1] [​2n * (2n − 1)] / [n * (n − 1)] = [12 / 1]

2n − 1 = 6n − 6

4n = 5

n = 5 / 4 which is not possible.

So, the right question is

2nC3 : nC2 = 12 : 1

ⁿCₐ = n! / a! (n – a)!

[2n (2n – 1) (2n – 2) (2n – 3)!] / [3! (2n – 3)!] : [n (n – 1) (n – 2)!] / [2! (n – 2)!] = [12 : 1]

{2n (2n – 1) (2n – 1) / 6} / {n (n – 1) / 2} = 12 / 1

{4n (2n – 1) (n – 1)} / 3 {n (n – 1)} = 12

4 (2n – 1) / 3 = 12

2n – 1 = 9

2n = 10

n = 5

(ii) 2nC3 : nC3 = 11 : 1

[{2n (2n – 1) (2n – 2) (2n – 3)!] / [3! (2n – 3)!}] / [{n (n – 1) (n – 2) (n – 3)!] / [3! (n – 3)!}] = [11 / 1]

{4n (n – 1) (2n – 1) / 6} / {n (n – 1) (n – 2) / 6} = 11

4 (2n -1) / (n – 2) = 11

8n – 4 = 11n – 22

8n – 11n = -22 + 4

-3n = -18

n = 6

Question 13: Find the number of chords passing through 11 points on the circumference of a circle.

Solution:

The number of the points on the circumference of circle = 11

A chord is formed by joining two points.

Hence taking 2 points out of 11 points number of chords = 11C2

= 11! / 2! (11 – 2)!

= 11! / 2! 9!

= [11 * 10 * 9!] / 2! 9!

= [11 * 10] / [2!]

= 110 / 2

= 55

Question 14: There are n points in a plane in which m points are collinear. How many triangles will be formed by joining three points?

Solution:

For making a triangle we need three points, thus if 3 points out of n points are not in a line, then nC3 triangles can be formed by n points.

But point m is on a line, thus mC3 triangle formed is less.

Hence, the required number of triangles = nC3mC3.

Question 15: Find the number of diagonals of a decagon.

Solution:

There are 10 vertices in a decagon. Now by joining two vertices, a side of a decagon or a diagonal is obtained.

Hence, the number of the line segments obtained by joining of vertices of decagon

= number of ways taking 2 points out of 10 points

= 10C2

Required number of diagonals

= {[10 (10 – 1)] / } – 10

= {[10 * (10 – 3)]} / 

= 70 / 2

= 35

Hence, the number of diagonals is 35.

Question 16: There are 5 empty seats in a train, then in how many ways three travellers can sit on these seats?

Solution:

A number of ways of seating of 3 travellers out of 5 seats.

= 5P3

= 5! / [5 – 3]!

= [5 * 4 * 3 * 2!] / [2!]

= 5 * 4 * 3

= 60

Hence, 3 travellers can be seated in 60 ways.

Question 17: A group of 7 has to be formed from 6 boys and 4 girls. In how many ways a group can be formed if boys are in the majority in this group?

Solution:

Number of boys = 6

Number of girls = 4

Number of members in the group = 7

On putting boys in the majority in the group, the groups has to be chosen in the following ways:

4 boys + 3 girls

5 boys + 2 girls

6 boys + 1 girl

Number of ways choosing 4 boys and 3 girls = 6C4 × 4C3

Number of ways choosing 5 boys and 2 girls = 6C5 × 4C2

Number of ways choosing 6 boys and 1 girl = 6C6 × 4C1

Required number of ways forming the group

= 6C4 × 4C3 + 6C5 × 4C2 + 6C6 × 4C1

= 15 * 4 + 6 * 6 + 1 * 4

= 60 + 36 + 4

= 100

Hence, the group can be formed in 100 ways.

Question 18: In the conference of 8 persons, every person handshakes with each other only once, then find the total number of hands shook.

Solution:

When two persons shake hands with each other, then this is taken as one handshake.

Hence, the number of handshakes is equal to choosing 2 persons out of 8 persons.

Number of handshakes = 8C2

= 8! / 2! [8 – 2]!

= [8 * 7 * 6!] / [2 * 6!]

= 56 / 2

= 28

Hence, the required number of hands-shaken is 28.

Question 19: In how many ways 6 men and 6 women can sit around the round table whereas any two women never sit together?

Solution:

First, the men sit in an order such that there is a place vacant between two men.

Then, the number of ways of 6 men sitting = (6 – 1)! = 5!

The number of ways of 6 women sitting in vacant places = 6!

Number of permutations = (6 – 1)! = 5!

The number of ways to seat 6 women in the vacant place = 6!

The number of permutations = 5! × 6!

= (5 * 4 * 3 * 2 * 1) * (6 * 5 * 4 * 3 * 2 * 1)

= 86400