Permutation and Combination for Class 11 Notes
Permutation and Combination class 11 is one of the most important topics for the students. In this chapter, important topics like permutation, combination, and the relationship between permutation and combination are covered. Also, examples of both permutation and combination for class 11 are given for students’ reference. This topic is not only important for school studies, but it is also important for other competitive examinations. So, it is mandatory to get the information on Maths class 11 tricks, permutation and combination formulas also.
Permutations and Combinations Class 11 Chapter 7 Concepts
- Fundamental Principle of Counting
Permutations when all the objects are distinct
Permutations when all the objects are not distinct
Fundamental Principle of Counting
“If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m × n.”
The above principle can be generalised for any finite number of events.
“If an event can occur in m different ways, following which another event can occur in n different ways, following which a third event can occur in p different ways, then the total number of occurrence to ‘the events in the given order is m × n × p.”
Learn more about the fundamental principle of counting here.
A permutation is defined as an arrangement in a definite order of a number of objects taken, some or all at a time. Counting permutations are merely counting the number of ways in which some or all objects at a time are rearranged. The convenient expression to denote permutation is defined as “ nPr ”.
The permutation formula is given by,
|Pr = n!/(n-r)! ; 0 ≤ r ≤ n|
Where the symbol “!” denotes the factorial which means that the product of all the integers is less than or equal to n but it should be greater than or equal to 1.
The notation n! represents the product of first n natural numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this symbol as ‘n factorial’.
Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !
1! = 1
2! = 1 x 2 = 2
3! = 1 x 2 x 3 = 6
4! = 1 x 2 x 3 x 4 = 24, which are the factors of the given number.
Click here to know more about factorial.
Permutation When all the Objects are Distinct
There are some theorems involved in finding the permutations when all the objects are distinct. They are :
Theorem 1: If the number of permutations of n different objects taken r at a time, it will satisfy the condition 0 < r ≤ n and the objects which do not repeat is n ( n – 1) ( n – 2)……( n – r + 1), then the notation to denote the permutation is given by “ n Pr”
Theorem 2: The number of permutations of different objects “n” taken r at a time, where repetition is allowed and is given by nr .
Permutation When all the Objects are not Distinct Objects
Theorem 3: To find the number of permutations of the objects ‘n’, and ‘p’s are of the objects of the same kind and rest is all different is given as n! / p!
Theorem 4: The number of permutations of n objects, where p1 are the objects of one kind, p2 are of the second kind, …, pk is of the kth kind and the rest, if any, are of a different kind, then the permutation is given by n! / ( p1!p2!…Pk!)
To get more knowledge about permutations, visit here.
The combination is a selection of a part of a set of objects or a selection of all objects when the order doesn’t matter. Therefore, the number of combinations of n objects taken r at a time and the combination formula is given by;
|nCr = n(n-1)(n-2)…(n-r+1)/ r! = n!/ r!(n-r)!= nPr /r!|
To understand more about combinations, visit here.
Relationship Between Permutation and combination
In permutation and combination for class 11, the relationship between the two concepts is given by two theorems. They are;
Theorem 5: nPr = nCr r! ; if 0 < r ≤ n.
Theorem 6: nCr + nCr-1 = n+1Cr
Problems based on Permutations and Combinations
The sample for permutation and combination for class 11 is given here. Go through the given example to get a clear idea.
Question 1: Find the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
(i) four cards are of the same suit,
(ii) four cards belong to four different suits,
(iii) are face cards,
(iv) two are red and two are black cards,
(v) cards are of the same colour?
There will be a number of possible ways for choosing 4 cards from 52 cards as there are combinations of 52 different things when we take 4 at a time.
Therefore, the required number of ways = 52C4
= 52! / (4! 48!) = (49x50x51x52) / (2x3x4)
(i) Four cards of the same suit:
There are four suits: Spade, heart, Club, diamond. Totally, there are 13 cards of each suit
Therefore, the required number of ways are given by 13C4 + 13C4 + 13C4 + 13C4
= 4(13! / (4! 9! )) = 2860
(ii) four cards belong to four different suits:
Since there are 13 cards in each suit. Therefore choosing 1 card from 13 cards of each suit, it becomes
= 13C1 + 13C1 + 13C1 + 13C1 = 134
(iii) Face cards :
There are 12 face cards and 4 cards are selected from these 12 cards, it becomes
Therefore, the required number of ways = 12! / ( 4! 8!) = 495
(iv) Two red cards and two black cards:
There are 26 red and 26 black cards in a pack of52 cards.
Therefore, the required number of ways = 26C2 x 26C2
(v) Cards of the same color:
Out of 26 red cards and 26 black cards, 4 red and black cards are selected in 26C4 ways. So, the required number of ways = 26C4 + 26C4
= 2 (26! / 4! 22! )
Question 2: How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Digits: 1, 2, 3, 4, 5, 6, 7, 8, 9
We can form several four-digit numbers using these 9 digits. Suppose 3456 and 6543 are two 4-digit numbers. Here, the order of digits is not a matter. Thus, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
9P4 = 9!/(9 – 4)!
= (9 × 8 × 7 × 6 × 5!)/5!
Hence, 3024 four-digit numbers can be formed using the digits from 1 to 9 without repetition.
Question 3: How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but the first letter is a vowel?
Given word: MONDAY
(i) Number of letters to be used = 4
Number of permutations = 6P4 = 6!/(6 – 4)! = 6!/2! = 720/2 = 360
Therefore, we can form 350 words with 4 letters from the word MONDAY.
(ii) Number of letters to be used = 6
Number of permutations = 6P6 = 6!/(6 – 6)! = 6!/0! = 720/1 = 720
Therefore, we can form 720 words using all the letters from the word MONDAY.
(iii) Number of vowels in the word MONDAY = 2 (O and A)
Number of ways that the first letter is a vowel = 2P1 = 2!/(2 – 1)! = 2!/1! = 2
Now, the remaining places = 5
Remaining letters = 5
These can be arranged in 5! ways, i.e. 120
Therefore, the total number of words can be formed with the first letter as vowel = 2 × 120 = 240.
- In an examination, in how many ways can a student select the questions if the question paper consists of 25 questions divided into Part I and Part II, containing 10 and 15 questions, respectively. A student is required to attempt 20 questions in all, selecting at least 5 from each part?
- If each selection of 4 cards has exactly one queen in a deck of 52 cards. Determine the number of 4-card combinations out of them
- How many such arrangements are possible if we need to seat 6 men and 5 women in a row so that the women occupy the even places.
- In how many ways can the outing party be chosen for a class of 40 students, if 25 are to be chosen.
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