# 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

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 = $$\frac { 6! }{ 2! }$$
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