Prepare the important questions for class 11 Maths Chapter 7 – Permutations and Combinations, which is given here. Students can learn how to find the factorial of a number, and how to compute the permutations and combinations for the given condition. This chapter concerns with computing the number of different ways of the arrangement of a given number of objects without listing the objects. Also, prepare all the chapters important questions for class 11 Maths here.
Class 11 Maths Chapter 7 – permutations and combinations describe the following important concepts:
- Fundamental Principle of Counting
- Multiplication Principle
- Addition Principle
- Permutation of n different objects
- When the repetition of an object is allowed
- Permutations when the objects are not distinct
- Important 1 Mark Questions for CBSE Class 11 Maths
- Important 4 Marks Questions for CBSE Class 11 Maths
- Important 6 Marks Questions for CBSE Class 11 Maths
Class 11 Chapter 7 – Permutations and Combinations Important Questions with Solutions
Practice the given important questions for class 11 Maths Chapter 7 – permutations and combinations given below.
Find the 3-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5 assuming that
a) digits can be repeated.
b) digits are not allowed to be repeated.
a) By the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125
b) By the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60
A coin is tossed 6 times, and the outcomes are noted. How many possible outcomes can be there?
When we toss a coin once, the number of outcomes we get is 2 (Either Head or tail)
So, in each throw, the no. of ways to get a different face will be 2.
Therefore, by the multiplication principle, the required no. of possible outcomes is
2 x 2 x 2 x 2 × 2 × 2 = 64
Evaluate the following
(i) 6 ! (ii) 5 ! – 2 !
(i) 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720
(ii) 5! = 1 × 2 × 3 × 4 x 5 = 120
As 2! = 1 × 2 = 2
Therefore, 5 ! – 2 ! = 120-2 = 118.
From a team of 6 students, in how many ways can we choose a captain and vice-captain assuming one person can not hold more than one position?
From a team of 6 students, two students are to be chosen in such a way that one student will hold only one position.
Here, the no. of ways of choosing a captain and vice-captain is the permutation of 6 different things taken 2 at a time.
So, 6P2 = 6! / ( 6 -2 )! = 6! / 4! = 30
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter Exactly once?
Number of letters in word EQUATION` = 8
n = 8
If all letters of the word used at a time
r = 8
Different numbers formed = nPr
= 8!/(8 8)!
= 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word – DAUGHTER?
No. of Vowels in the word – DAUGHTER is 3.
No. of Consonants in the word Daughter is 5.
No of ways to select a vowel = 3c2 = 3!/2!(3 – 2)! = 3
No. of ways to select a consonant = 5c3 = 5!/3!(5 – 3)! = 10
Now you know that the number of combinations of 3 consonants and 2 vowels = 10x 3 = 30
Total number of words = 30 x 5! = 3600 ways.
It is needed to seat 5 boys and 4 girls in a row so that the girl gets the even places. How many are such arrangements possible?
5 boys and 4 girls are to be seated in a row so that the girl gets the even places.
The 5 boys can be seated in 5! Ways.
For each of the arrangement, the 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).
B x B x B x B x B
So, the girls can be seated in 4! Ways.
Hence, the possible number of arrangements = 4! × 5! = 24 × 120 = 2880
Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Take a deck of 52 cards,
To get exactly one king, 5-card combinations have to be made. It should be made in such a way that in each selection of 5 cards, or in a deck of 52 cards, there will be 4 kings.
To select 1 king out of 4 kings = 4c1
To select 4 cards out of the remaining 48 cards = 48c4
To get the needed number of 5 card combination = 4c1 x 48c4
= 4x2x 47x 46×45
= 778320 ways.
Find the number of 6 digit numbers that can be formed by using the digits 0, 1, 3, 5, 7, and 9. These digits shall be divisible by 10, and no digit shall be repeated?
The number which has a 0 in its units place is divisible by 10.
If we put 0 in the unit place, _ _ _ _ 0, there will be as many ways to fill 5 vacant places. (1, 3, 5, 7, 9)
The five vacant places can be filled in 5! ways = 120.
Evaluate: 10! – 6!
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x1 = 3628800
6! = 6 X 5 x 4 x 3 x 2 x 1 = 720
10! – 6! = 3628800 – 720 = 3628080
Practice Problems for Class 11 Maths Chapter 7
Solve the problems from class 11 Maths Chapter 7 – permutations and combinations:
- Calculate how many numbers are there between 99 and 1000 having atleast one of their digits 7?
- A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Mathematics Part II, unless Mathematics Part I is also borrowed. In how many ways can he choose the three books to be borrowed?
- Given that 5 flags are of different colours. Calculate how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
- Compute n!/ r!(n-r)!, when n=5 and r=2.
- Determine the number of permutations of the letters of the word ALLAHABAD.
- How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?
- Find the number of different words that can be formed from the letters of the word TRIANGLE, so that no vowels are together.
- A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
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