Class 11 : Permutation & Combination : Formulas

 

Fundamental Principles of Counting

1. Multiplication Principle

If first operation can be performed in m ways and then a second operation can be performed in n ways. Then, the two operations taken together can be performed in mn ways. This can be extended to any finite number of operations.

2. Addition Principle

If first operation can be performed in m ways and another operation, which is independent of the first, can be performed in n ways. Then, either of the two operations can be performed in m + n ways. This can be extended to any finite number of exclusive events.

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Factorial

For any natural number n, we define factorial as n ! or n = n(n – 1)(n – 2) … 3 x 2 x 1 and 0!= 1!= 1

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.

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Permutation

Each of the different arrangement which can be made by taking some or all of a number of things is called a permutation.

Mathematically  The number of ways of arranging n distinct objects in a row taking r (0 ≤ r ≤ n) at a time is denoted by P(n ,r) or ⁿp_r

NOTE : 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,

nPr = n!/(n-r)! ; 0 ≤ r ≤ n

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’ 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!)

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Combination

Each of the different groups or selections which can be made by some or all of a number of given things without reference to the order of the
things in each group is called a combination.

Mathematically The number of combinations of n different things taken r at a time is

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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

Sample Question : Find the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these

1. four cards are of the same suit,
2. four cards belong to four different suits,
3. are face cards,
4. two are red and two are black cards,
5. 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 = 52C4

= 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 13C4 + 13C4 + 13C4 + 13C4

= 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

= 13C1 + 13C1 + 13C1 + 13C1 = 134

(3)Face cards :

There are 12 face cards and 4 cards are selected from these 12 cards, it becomes

= 12C4

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 = 26C2 x 26C2

=(26!2!24!)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 26C4 ways. So, the required number of ways = 26C4 + 26C4

= 2 (26! / 4! 22! )

=29900.

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