Xi ‘a’

V.MURUGAN

FUNDAMENTAL PRINCIPAL OF

COUNTING

FUNDAMENTAL PRINCIPAL OF COUNTING MULTIPLICATION PRINCIPLE If first operation can be done by m ways & second

operation can be done by n ways Then total no of ways by which both operation can be done

simultaneously =m x n

ADDITION PRINCIPLE If a certain operation can be performed in m ways and

another operation can be performed in n ways then the total number of ways in witch either of the two operation can be performed is

m + n.

How many 3 digit no can be formed by using digits 8,9,2,7 without repeating any digit?

How many are greater than 800 ?A three digit number has three places to be filled

Now hunderd’th place can be filled by 4 ways ,After this tenth place can be filled by 3 waysAfter this unit place can be filled by 2 waysTotal 3 digits no we can form =4x3x2= 24

Hundred place

Tenth place

Unit place

EXAMPLE :-

SECOND PARTTo find total number greater than 800 (by digits

8,9,2,7 )

(we observe that numbers like 827 , 972 etc. starting with either 8 or by 9 are greater than 800 in this case)

Hence Hundred th place can be filled by 2 ways (by 8 or 9)After this tenth place can be filled by 3 ways After this unit place can be filled by 2 ways Total 3 digits no greater than 800 are =2x3x2=12

Hundred place

Tenth place

Unit place

8 9 2 7

Both are ways to count the possibilitiesThe difference between them is whether

order matters or notConsider a poker hand:

A♦, 5♥, 7♣, 10♠, K♠Is that the same hand as:

K♠, 10♠, 7♣, 5♥, A♦Does the order the cards are handed out

matter?If yes, then we are dealing with permutationsIf no, then we are dealing with combinations

Permutations vs. Combinations

A permutation of given objects is an arrangements of that objects in a specific order.

Suppose we have three objects A,B,C.

so there are 6 different permutations (or

arrangements ) In PERMUTATATION order

of objects important . ABC ≠ ACB

A CB

A

A

A

A

A

B

B

B

B

B

C

CC

C

C

Permutations

PERMUTATION OF DISTINCT OBJECTS The total number of different permutation of n

distinct objects taken r at a time without repetition is denoted by nPr and given by

n Pr = where n!=

1x2x3x. . .xn Example Suppose we have 7 distinct objects and out

of it we have to take 3 and arrange Then total number of possible arrangements would

be

7P3 = = 840

Where 7!= 7x6x5x4x3x2x1

Suppose there are n objects and we have to arrange all these objects taken all at the same time

Then total number of such arrangements ORTotal number of Permutation will be = n Pn

=

=

= n!NotationInstead of writing the whole formula,

people use different notations such as these:

The factorial function (symbol: !) just means to multiply a series of descending natural numbers.

Examples: 4! = 4 × 3 × 2 × 1 = 24 7! = 7 × 6 × 5 × 4 × 3 ×

2 × 1 = 5040 1! = 1

Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations.

Q(1) In how many ways 2 Gents and 6 Ladies can sit in a row for a photograph if Gents are to occupy extreme positions ?

SOLUTION

Here 2 Gents can sit by =2! Ways ( As they can interchange there positions so first

operation can be done by 2! Ways)After this 6 Ladies can sit by =6! Ways (Ladies can interchange their positions among

themselves so second operation can be done by 6! Ways )

Hence total number of possible ways are = 2!x6! =1440

L L L L L L GG

EXAMPLE :-

In how many ways 3 boys and 5 girls sit in a row so that no two boys are together ?

Girls can sit by 5! Ways After this now out of 6 possible places for boys to sit

3 boys can sit by 6P3 ways

Hence total number of ways = 5!x 6P3

G G G G G

A combination is selection of objects in which order is immaterial

Suppose out of 15 girls a team of 3 girls is to select for Rangoli competition

Here it does not matter if a particular girl is selected in team in first selection or in second or in third .

Here only it matter whether she is in team or not

i. e. order of selection does not matter .In Permutation : Ordered SelectionIn combination : Selection ( Order does not

matter)

COMBINATION

SUPPOSE 3 OBJECTS A B C ARE THERE We have to select 2 objects to form a team Then possible selection ( or possible team )AB ,AC,BC i.e. 3 different team can be formedRemark : Note that here team AB and BA is same

OBJECTS A, B,C

COMBINATIONS

AB,BC,CA

PERMUTATIONS AB,BA,BC,CB,AC,

CA

COMBINATION OF DISTINCT OBJECTS A combination of n distinct objects taken r at a time is

a selection of r objects out of these n objects ( 0 ≤ r ≤ n).

Then the total number of different combinations of n distinct objects taken r at a time without repetition is denoted by n Cr and given by

n Cr = Suppose we have 7 distinct objects and out of it we

have to select 3 to form a team .Then total number of possible selection would be

7C3 = = = = 35

In a box there are 7 pens and 5 pencils . If any 4

items are to be selected from these Find in how many ways we can selectA) exactly 3 pens B) no pen C) at least one pen D) at most two pens Solution :-A) 7C3 x 5C1

B) 5C4

C) either 1 pen OR 2 pens OR 3 pens OR 4 pens

7C1 x 5C3 + 7C2 x 5C2 + 7C3 x 5C1 + 7C4

D) either no pen OR 1 pens OR 2 pens 7C0 x 5C4 + 7C1 x 5C3 + 7C2 x 5C2

EXAMPLE:-