Permutation and Combination class 11 is one of the most important topics for the students. In this chapter, the important topics like permutation, combination, and the relationship between permutation and combination is covered. Also, the examples of both permutation and combination class 11 notes are given for student’s 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 permutation and combination class 11 tricks, permutation formula and combination formula also.
Permutation
Permutation Definition: 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 “ ^{n}P_{r }”.
The permutation formula is given by,
\(P_{r}=\frac{n!}{(n-r)!};0\leq r\leq n\)Where the symbol “!” denotes the factorial which means that the product of all the integer less than or equal to n but it should be greater than or equal to 1.
For example,
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.
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 }P_{r}”
Theorem 2: The number of permutations of different objects “n” taken r at a time, where repetition is allowed and is given by n^{r} .
Permutation When all the Objects are not Distinct Objects
Theorem 3: To find the number of permutations of the objects ‘n’, and ‘p’ 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 p_{1} are the objects of one kind, p_{2} are of the second kind, …, p_{k }is of the k^{th} kind and the rest, if any, are of a different kind then the permutation is given by n! / ( p_{1}!p_{2}!…P_{k}!)
Combination
Combination Definition: The combination is a selection of a part of a set of objects or 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
_{n}C_{r} = n(n-1)(n-2)…(n-r+1)/ r! = n!/ r!(n-r)! = _{n}P_{r} / r!
Relationship Between Permutation and combination
In permutation and combination class 11, the relationship between the two concepts is given by two theorems. They are
Theorem 5: ^{n}P_{r} = ^{n}C_{r } r! ; if 0 < r ≤ n.
Theorem 6: ^{n}C_{r }+ ^{n}C_{r-1} = ^{n+1}C_{r}
Permutation and Combination Questions
The sample for permutation and combination class 11 is given here. Go through the given example to get a clear idea.
Question :
Find the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
- four cards are of the same suit,
- four cards belong to four different suits,
- are face cards,
- two are red and two are black cards,
- cards are of the same colour?
Solution :
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 = ^{52}C_{4 }
= 52! / (4! 48!) = (49x50x51x52) / (2x3x4)
= 270725
(1) 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 ^{13}C_{4 }+ ^{13}C_{4 }+ ^{13}C_{4 }+ ^{13}C_{4}
= 4(13! / (4! 9! )) = 2860
(2) 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
= ^{13}C_{1 }+ ^{13}C_{1 }+ ^{13}C_{1 }+ ^{13}C_{1 } = 13^{4}
(3)Face cards :
There are 12 face cards and 4 cards are selected from these 12 cards, it becomes
= ^{12}C_{4 }
Therefore, the required number of ways = 12! / ( 4! 8!) = 495
(4) 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 = ^{26}C_{2 } x ^{26}C_{2 }
=\(\left ( \frac{26!}{2!24!} \right )^{2}\)
= (325)^{2}
=105625
(5) Cards of the same color:
Out of 26 red cards and 26 black cards, 4 red and black cards are selected in ^{26}C_{4} ways. So, the required number of ways = ^{26}C_{4} + ^{26}C_{4}
= 2 (26! / 4! 22! )
=29900.
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