4. 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.

5.  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
Unit
Hundred Tenth
place
place place
 Now hunderd’th place can be filled by 4 ways ,
 After this tenth place can be filled by 3 ways
 After this unit place can be filled by 2 ways
 Total 3 digits no we can form =4x3x2= 24

6.  SECOND PART
 To find total number greater than 800 (by digits 8,9,2,7 )
Hundred Tenth Unit
place place place
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

7.  Both are ways to count the possibilities
 The difference between them is whether order
matters or not
 Consider 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 permutations
 If no, then we are dealing with combinations

9.  A permutation of given objects is an arrangements of that
objects in a specific order.
 Suppose we have three objects A,B,C.
A B C C A B
A C B C B A
B A C
so there are 6 different permutations
(or
B C A
arrangements )
In PERMUTATATION order of objects
important . ABC ≠ ACB

10.  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
 nP = where n!= 1x2x3x. . .xn
r

 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

11.  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
 OR
 Total number of Permutation will be = nP
n
=
=
= n!
 Notation
 Instead of writing the whole formula, people use
different notations such as these:

12. 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.

13.  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
G L L L L L L G
 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

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

16.  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 Selection
 In combination : Selection ( Order does not
matter)

17. 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 formed
Remark : Note that here team AB and BA is same
OBJECTS A, B,C
COMBINATIONS PERMUTATIONS
AB,BC,CA AB,BA,BC,CB,AC,CA

18.  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
 nC
r =

 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


19.  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 select
 A) 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
 7C
1 x 5C3 + 7C2 x 5C2 + 7C3 x 5C1 + 7C4
 D) either no pen OR 1 pens OR 2 pens
 7C x 5C4 + 7C1 x 5C3 + 7C2 x 5C2
0