Class 10 Advanced Mathematics Solution Chapter 6 Pdf
In this Unit we will cover the Permutation and Combination, also we will discuss the solution of Class 10 Advanced Mathematics Solution of Chapter 6 Permutation and Combination Exercise-6.1. Here you will get the correct solution from our experts. so let's start with the introduction...
Class 10 Mathematics Solution
NCERT Class 10 Science Notes on Electricity
Introduction:
Suppose we have three objects A, B, C of different colors. Now, taking two of them at a time we get the following three units. AB, AC, BC.
The first unit AB can be arranged in two ways: AB and BA.
Arranging this way from these three different objects taking two at a time we get six arrangements viz. AB. BA, AC, CA, BC, and CB.
Each of the above six arrangements is called a Permutation. On either hand, each unit obtained here taking two at a time is called a Combination.
What is Permutation?
Permutation: The different kinds of arrangements that can be made out of a given set of different objects taking some or all of them at a time is called their permutation.
The permutation of n different objects taking 'r' at a time is symbolized as nPr or P(n, r).
What is Combination?
Combination: From a collection of different objects, the different kinds of selections that can be made out of taking some or all of them are called their combination.
Also Check: CLASS 10 GENERAL MATHS IMPORTANT QUESTIONS PDF
The combination of n different objects taking 'r' of them at a time is symbolized as nCr or C(n, r)
CHAPTER-6
PERMUTATION AND COMBINATION
EXERCISE-6.1
( Question with Solution)
1.Find the value.(i) 5!
= 5.4.3.2.1
=120
(ii) 4! + 5!
= 4! + 5!
= 4! + 5.4!
= 4! (1+5)
= 6. 4!
= 6.4.3.2.1
=144
(iii) 9! / 7!
= (9.8.7!)/7!
= 9.8
=72
(iv) 6! -3!
= 6.5.4.3.2.1 - 3.2.1
=720 - 6
= 714
(v) 12!/ 2! 10!
= (12.11. 10!)/ 2. 10!
= 6.11
= 66
(vi) 3! 7!
= 3.2.1.7.6.5.4.3.2.1
= 30240
2.Show that, 2n! = 2n. n! {1.3.5…………..(2n-1)}
Solutions: LHS: 2n!
= 2n (2n-1)(2n-2)(2n-3)……….4.3.1
={2n (2n-2)(2n-4)……4.2}{(2n-1)…………3.1}
= 2n { n ( n-1) (n-2)…..2.1}( 2n-1) (2n-3)……..3.1}
= 2n 2n! { 1.3.5………… (2n - 3) ( 2n - 1)}
= R.H.S
Hence Showed.
Solve applying the fundamental principle of counting:
3. How many words ( the words may or may not have meaning) can be formed with the letters of the word ENGLISH taking three at a time?
Solution: There are 7 letters in the word ENGLISH
The first place of the word can be select in 7 ways. After selecting the first place by 7 letters the remaining 6 places will be fillup by 6 letters and the 3rd place will be fillup by the remaining 5 letters in 5 ways.
∴ By the fundamental principle of counting the total number of words = 7ｘ 6ｘ 5 = 210
4. If repetition is allowed, how many even numbers of
two digits can be formed with the digits 1, 2, 3, 4, 5?
Solution: To make the number even its unit place digit must be
either 2 or 4. The unit place can be fillup by 2 ways and the tens places can be
fillup by 5 ways. Since repetition of digits is allowed therefore in tens
place we can put 1, 2, 3, 4, and 5 also.
∴ By the fundamental principle of counting the total number of two digits even number
= 2ｘ 5 = 10
5. Without repetition how many numbers of 5 digits can
be formed with the digits 0, 1, 2, 3, 5, 6, 7, 8, 9 ?
Solution: Without repetition of digits 0, 1, 2, 3, 5, 6, 7, 8, 9 numbers
of 5 digits can be formed
= 9ｘ 9ｘ 8 ｘ7ｘ6 = 27216
6. Without repetition, how many 4 digits numbers can be formed with the digits 1, 3, 5, 7, 9?
Solution: Without repetition of digits 1, 3, 5, 7, 9 numbers of 4 digits can be a form
= 5ｘ 4ｘ 3ｘ2= 120
7. How many ways the letters of the word NUMBER can be arranged?
(I) How many of these will start with M?
(ii) How many of these do not end with B?
Solutions: The total number of letters in the word NUMBER is 6
∴ By the fundamental principle of counting the total number of arrangement of the letters of te word NUMBER = 6 ｘ5 ｘ4 ｘ3 ｘ2 ｘ1 = 720
(i) Words start with M ie leaving M by the remaining 5 letters we have to fill up.
∴ Total numbers of words started with M = 5 ｘ4 ｘ3 ｘ2 ｘ1 = 120
(ii) Total number of words do not end with b = 5 ｘ5 ｘ4 ｘ3 ｘ2 = 600
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