RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1

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

Question 1.
Find the value of n, whereas n
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 1
Solution:
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 2
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 3
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 4

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
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 5
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.
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 6
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
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 7

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.
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 8
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.
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 9

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.
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 10
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 = \(\frac { 6! }{ 2! }\)
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
RBSE Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Ex 6.1 11

RBSE Solutions for Class 11 Maths