Basic Concepts of Permutations and Combinations Solutions CA maths chapter 5 exercise 5A
CA foundation maths solutions Chapter 5 Basic Concepts of Permutations and Combinations Exercise 5A 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
CA maths solutions chapter 5 Basic Concepts of Permutations and Combinations
CA foundation maths solutions
Chaper 5
Basic Concepts of Permutations and Combinations
Exercise 5B
3. 7! Is equal to
Solution:
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
4. 0! is a symbol equal to
Solution:
From the theory
0! = 1
Basic Concepts of Permutations and Combinations CA foundation maths solution Exercise 5A
16. 10 examination papers are arranged in such a way that the best and worst papers never come together. The number of arrangements is
Solution:
We will make one group of best and worst papers remaining eight individual papers are considered as 8 different groups.
There are total 9 groups has to arrange it can be done in 9! different ways.
Two papers best and worst can be arrange with in 2! different ways.
The number of arrangements of these two papers together = 9! × 2!
The number of arrangements of such two best and worst paper can never be together =
= (10 – 2)! × 9! = 8! × 9!
17. n articles are arranged in such a way that 2 particular articles never come together. The number of such arrangements is
Solution:
We will make one group of 2 articles. Remaining (n- 2) individuals articles are considered as (n – 2) different groups.
The total (n -2) + 1 = (n – 1) groups, they has to arrange in (n – 1)! different ways.
These 2 particular articles can arrange with in themselves in 2! different ways.
The number of arrangements of these two particular article together = (n – 1)! + 2!
The number of arrangement of 2 particular articles can never be together
= ni – 2 × ( n – 1)! = n × (n-1)! -2×(n-1)
= (n – 2) × (n – 1)!
18. If, 12 school teams are participating in a quiz contest then the number of ways the first, second and third positions may be won is
19. The sum of all four digit number containing the digits 2, 4, 6, 8 without repetition is
Solution:
20. The number of 4 digit numbers greater than 5,000 can be formed out of the digits 3, 4, 5, 6 and 7 ( No. digit is repeated). The number of such is
21. 4 digit number to be formed out of the figures 0, 1, 2, 3, 4 (no digit is repeated) then number of such numbers is
We cannot select 0.
22. The number of ways the letters of the word ‘TRIANGLE’ to be arranged so that the word ‘angle’ will be always present is.
Solution
‘The word ‘angle’ as a group. The remaining letters are considered as a 3 groups.
It can be done in 4! different ways
The number of words formed = 4!= 4×3×2 = 24
23. If the letters word ‘DAUGHTER’ are to be arranged so that vowels occupy the odd places, then number of different words are
Solution:
The word ‘DAUGHTER’ . It has 4 position and 3 vovels.
Total positions are 5.
The arrangement can be done in 5! different ways
The number of word formed
Note: You can observe the solutions. You will try them in your own method.