RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

Rajasthan Board RBSE Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1

Question 1.
Find the value of n, whereas n

Solution:

Question 2.
Find the number of different words formed by letters of word ALLAHABAD.
Solution:
There are 9 letters in given word, in which 4A, 2L and other are different.
So, by taking all letters, number of different words are

Hence, number of words are 7560.

Question 3.
How many words can be formed using letters of 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 starting and E at the end, then the number of remaining letters = 6
Hence, 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 has 3 on hundredth’s place and 5 on one’s place and the total number of digits is 4.

Hence, total 12 number 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 numbers becomes of 5-digit number then the 5-digit number formed is = 5! = 5 × 4 × 3 × 2 × 1 = 120
Hence, the numbers that do not have zero in the begining 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 is repeated?
Solution:
Given number of digits = 6
For forming 3-digit number, we have to choose of the three digit from these 6-digits.
Hence, total numbers which have 3 digits

Question 7.
How many ways 15 members of a committee can sit around the round table whereas secretary sit on one side and vice secretary on other side of Director ?
Solution:
Total number of members = 15
Here, the position of secretary, vice secretary and director are fixed but there is no condition of clockwise and anti clockwise.
Hence, total number of sitting ways according to condition = 2!
Now, we can change the position of 12 members only.

Hence, total number sitting ways of 12 members = 12!
Hence, total number of ways 15 member sit according to given condition = 12! × 2!

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 other station of this line?
Solution:
Total number of stations = 15
A person buys a ticket on a station then we can buy tickets of the remaining 14 stations.
Hence, a total number of methods of printing 14 tickets out of 15 tickets.

One ticket can be printed in 2 ways in this way total 14 tickets will be printed.
Hence, the total number of ways tickets be printed for 15 stations are = 15 × 14 = 210.

Question 9.
How many ways 10 different beads can be threads for making a garland whereas 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.

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 is divisible by 5 has 6 on thousand’s place and 0 or 5 on one’s place and the number is 4 digits number. As digits can be repeated.

Hence,
Total numbers = 1 × 10 × 10 × 2 = 200

Question 11.
How many words can be formed from letters of word SCHOOL, whereas both O’s not come together?
Solution:
There are total of 6 letters in word SCHOOL.
In it there are 2 O’s and other letters are different than the number of total permutations = 6!/2!
If 2 O’s take together then it will take as one letter then the number of permutations taking 2 O’s together = 5!
Hence, the number of permutations when 2 O’s are not together are

Rajasthan Board RBSE Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.2

Question 1.
Find the value of n, whereas:

Solution:

Question 2.
Find the value of ⁵⁰C₁₁ + ⁵⁰C₁₂ + ⁵¹C₁₃ – ⁵²C₁₃.
Solution.

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:
Total number of points on side AB, BC and CA = 3 + 4 + 5 = 12
Hence, number of triangles formed by 12 points


Points on sides AB, BC and CA are collinear.
Hence, any triangle cannot be formed by these points
Hence, total number of required triangles = 220 – 1 – 4 – 10 = 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:
Total number of balls in the box = 2 + 3 + 4 = 9
For at least one black ball we can choose it in the following ways.

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

Solution:
Total number of coloured flags = 6
Hence we can take any colour from 1 to 6.
Thus, the number of signals formed by using 6 different types of coloured flags

Question 6.
Number of diagonals in a polygon is 44 then find a number of its sides.
Solution:
Number of diagonals = 44
Formula : Number of diagonals = n(n−3)/2
where n = number of sides.
⇒ n(n−3)/2 = 44
⇒ n² – 3n = 88
⇒ n² – 3n – 88 = 0
On solving n = 11, -8
Negative number is not possible.
Hence, the number of sides = 11.

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
We Have to take 4 digits but digit 4 and 5 are compulsory.
Thus, we have to choose only 2 digits out of 6 – 2 = 4 digits.
Now, the numbers formed by using 4 digits is ⁴P₄.
Hence required number

Question 8.
In how many ways 6 ‘+’ and 4 ‘-‘ sign can be kept in a line whereas any two ‘-‘ occur together?
Solution:
First we put ‘+’ in a line. This can be done in 1 way because all ‘+’ are same.
+ + + + + +
Now we put ‘-‘ in the following way
– + – + – + – + – + – + –
Hence, the method of putting ‘-‘ sign in 4 out of 7 places = ⁷C₁ × 1 because ‘-‘ signs are same.
Total number of ways 6 ‘+’ and 4 ‘-‘ sign occur together

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:
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

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 this team can be formed?
Solution:
The method of choosing 2 out of 4 bowlers is ⁴C₂ and the method of choosing 9 out of remains to players ¹⁰C₉. Similarly, we can find methods for 3 and 4 bowlers.
Hence, total number of ways to select 11 players out of 14 players

Rajasthan Board RBSE Class 11 Maths Chapter 6 Permutations and Combinations Miscellaneous Exercise

Question 1.
If ⁿP_n-2 = 60, then n will be:
(A) 2
(B) 4
(C) 5
(D) 3
Solution:

Question 2.
ⁿP_r ÷ ⁿC_r is equal to:
(A) n!
(B) (n – r)!
(C) 1r!
(D) r!
Solution:

Hence, the option (D) is correct.

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.
Hence, option (B) is correct.

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!
Hence option (A) is correct.

Question 5.

Solution:

Question 6.
Find the value of ⁶¹C₅₇ – ⁶⁰C₅₆:
(A) ⁶¹C₅₈
(B) ⁶⁰C₅₇
(C) ⁶⁰C₅₈
(D) ⁶⁰C₅₆
Solution:

Question 7.
If ¹⁵C_3r = ¹⁵C_r+3, then r is equal to:
(A) 5
(B) 4
(C) 3
(D) 2
Solution

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

Hence, option (B) is correct.

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 are 6 and every time we take 6 letters.
Hence, required number of = ⌊6 = 6 × 5 × 4 × 3 × 2 × 1 = 720
Hence, option (D) is correct.

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.
Given: There are 4 points on the circumference of a circle.
Then, the number of required triangles

Hence, option (A) is correct.

Question 11.
If ⁿC₂ = ⁿC₇, then find ⁿC₁₆
Solution:

Question 12.
Find the value of n:
(i) ⁿC₂ : ⁿC₂ = 12 : 1
(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1
Solution:

Question 13.
Find the number of chords passing through 11 points on the circumference of a circle.
Solution:
Number of the point on the circumference of circle = 11
We know that a chord is formed by joining two points.
Hence taking 2 points out of 11 points number of chords = ¹¹C₂

Hence, the number of chords = 55.

Question 14.
Determine the number of combinations out of deck of 52 cards of each selection of 5 cards has exactly one ace.
Solution:
Number of cards in a deck = 52
Total number of aces = 4
Hence Number of remaining cards = 52 – 4 = 48
We have to make a collection of 5 cards, in which there is 1 ace and 4 other cards.
Hence, number of ways choosing 1 out of 4 ace.
Now,Number of ways choosing 4 cards out of 48 cards

Total number of collection having 5 cards
= ⁴C₁ × ⁴⁸C₄ = 4 × 194580 = 778320
Hence, number of combinations formed by 5 cards = 778320.

Question 15.
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 triangle can be formed by n points but m point is in a line, thus ^mC₃ triangle forms less.
Hence, required number of triangles = ⁿC₃ – ^mC₃.

Question 16.
Find the number of diagonals of a decagon.
Solution:
There are 10 vertices in a decagon. Now by joining two vertices, we get a side of decagon or a diagonal.
Hence, the number of the line segment by joining of vertices of decagon = number of ways taking 2 points out of 10 points

Hence, the number of diagonals is 35.

Question 17.
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.

Hence, 3 travellers can be seated in 60 ways.

Question 18.
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
One putting boys in the majority in the group
We have to choose in the following ways :
4 boys + 3 girls
5 boys + 2 girls
6 boys + 1 girls
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

Hence, the group can be formed in 100 ways.

Question 19.
In the conference of 8 persons, every person handshake with each other only once, then find the total number of hand shook.
Solution:
When two persons shake hand 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 handshake

Hence, the required number of hand shaken is 28.

Question 20.
In how many ways 6 men and 6 women can sit around the round table whereas any two women never sit together?
Solution:
First, we sit the men in an order that there is a place vacant between two men.
Then, number of ways sitting 6 men = (6 – 1)! = 5!
Now a number of ways sitting 6 women in the vacant places = 6!
Number of permutations = (6 – 1)! = 5!
Now, the number of ways sitting 6 women in the vacant place = 6!
Number of permutaions = 5! × 6!
= (5 × 4 × 3 × 2 × 1) × (6 × 5 × 4 × 3 × 2 × 1)
= 86400.

Question 21.
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S are together?
Solution:
Number of letters in word ASSASSINATION = 13
In this word A comes 3 times, S comes 4 times, I comes 2 times, N comes 2 times and other letters are different.
Now, taking 4’s together, then consider it as one letter, now adding other 9 letters and thus the permutation formed by 10 letters
= 10!/2!2!3!
[Because I comes 2 times, N comes 2 times and A comes 3 times]

Hence, the total number of ways is 151200.

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