Basic Concepts of Permutations and Combinations Exercise 5B Solutions CA maths chapter 5
CA foundation maths solutions Chapter 5 Basic Concepts of Permutations and Combinations Exercise 5B are given.
You should study the textbook lesson Basic Concepts of Permutations and Combinations very well.
You must practice all example Problems and solutions which are given in the textbook.
You can observe the solutions given below and try them in your own way.
Basic Concepts of Permutations and Combinations
Exercise 5A
Exercise 5B
Exercise 5C
Exercise 5D
CA maths solutions Exercise 5B chapter 5 Basic Concepts of Permutations and Combinations
CA foundation maths solutions
Chaper 5
Basic Concepts of Permutations and Combinations
Exercise 5B
Q. 1. The number of ways in which 7 girls form a ring is.
Solution:
One girl has to give the position.
Therefore, the number of ways in which 7 girls form a ring = 6! =6×5×4×3×2×1= 720
2. The number of ways in which 7 boys sit in a round table so that two particular boys may sit together is
Solution:
The two particular boys wants to sit together.
One boy has two possibility to select his position.
Then 5 positions are left there are five boys to occupy.
This correct arrangement can be done in 5! different ways.
The number of ways of arrangement of 7 boys around the table = 2 × 5! =2×5×4×3×2×1 = 240.
3. If 50 different jewels can be set to form a necklace then the number of ways is
Solution:
If we fix one jewel at a position. Then arrange the other 49 jewels.
The arrangement can be done in 49! ways.
They arrange in clockwise or anticlockwise direction. The order will be the same.
The number of arrangement of jewels
= 1/2×49!
4. Three ladies and three gents can be seated at a round table so that any two and only two of the ladies sit together. The number of ways is
Solution:
Total arrangement of three ladies under three gents = selection of two ladies × number of options for first lady × number of options for second lady × number of options for 3rd lady × number of options for 3 gents.
5. The number of ways in which the letters of the world ‘DOGMATIC’ can be arranged is
Solution:
The word ‘DOGMATIC’ has 8 alphabets.
The number of ways of arrangement of the word = 8! = 8×7×6×5×4×3×2×1 = 40320.
6. The numer of arrangements of 10 different things taken 4 at a time in which one particular thing always occurs is.
Solution;
Let w we have to rearrange four things but one thing should always occur. From remaining 9 things we will select 3 things.
7. The number of permutations of 10 different things taken 4 at a time in which one particular thing never occurs is
8. Mr. X and Mr Y enter into a railway compartment having six vacant seats. The number of ways in which they can occupy the seat is
9. The number of numbers lying between 100 and 1000 can be formed with the digits 1, 2, 3, 4. 5, 6, 7 is
10. The number of numbers lying between 10 and 1000 can be formed with the digits 2, 3, 4. 0, 8, 9 is
Solution:
We have to form the number using the digits 2, 3, 4, 0, 8, 9.
We can’t select 0. We can select any one digit out of 5 given digits 2,3,4,8,9.
It can be done in 5 different ways.
Number of two digit numbers form = 5×5 = 25
There are four digits. It can be done in 4 different ways.
Number of three digit numbers form =
5×5×4 = 100
The numbers lying between 10 and 100 = number of two digit number + number of three digit number = 25 + 100 = 125.
11. In a group of boys the number of arrangement of 4 boys is 12 times the number of arrangements of 2 boys. The number of boys in the group is
Basic Concepts of Permutations and Combinations CA foundation maths solution Exercise 5B
13. The total number of 9 digit numbers of different digits is
14. The number of ways in which 6 men can be arranged in a row so that they particular 3 men sit together, is
15. There are 5 speakers A, B, C, D and E. The number of ways in which A will speak always before B is.
16. There are 10 trains plying between Calcutta and Delhi The number of ways in which a person can go from Calcutta to Delhi and return by a different train is
Solution:
The person has 10 different options.
While coming back he has to select the different train.
Therefore he has 9 different options.
The number of ways he can travel to return
=10×9 = 90
17. The number of ways in which hate sweets of different sizes can be distributed among 8 persons of different ages so that they largest sweet always goes to be younger assuming that each one of them gets a sweet is
Solution:
Let we give the largest sweet to the youngest person.
We have 7 different sweets that have to be distributed among the remaining 7 persons.
It can be done by 7! ways.
The number of ways of distribution of sweets =7! = 7×6×5×4×3×2×1 = 5040.
18. The number of arrangements in which the letters of the world ‘MONDAY’be arranged so that the words thus formed begin with M and do not end with N is
Solution:
We select four alphabets O, D, A, Y
Four positions and four alphabets
The number of arrangements of the word ‘MONDAY’ = 1×4× 4! =4×24 = 96
19. The total number of ways in which six ‘+’ and four ‘-‘ signs can be arranged in a line such that no two ‘-‘ signs occur together is
Solution:
First we will arrange all + signs in a row.
It can be done in 6! ways.
Total number of arrangements of this = 6!/6! = 1
There are 7 positions. Among these four – signs.
20. The number of ways in which the letter of the word ‘MOBILE’ be arranged so that consonants always occupy the odd places is
21. 5 persons are sitting in a round table in such way that Tallest Person is always on the right side of the shortest person; the number of such arrangements is
The tallest person will sit to right side of the shortest person.
There are three chairs are vacant.
There are three persons. They can be arranged in 3! ways.
The total number of arrangements
= 1 × 1 × 3 = 6.
Note: You can observe the solutions. You will try them in your own method.