concept & examples on permutation & combination chapter.

1. PERMUTATION AND
COMBINATION

2. • 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.

3. • EXAMPLE suppose you want to get a policy to get tax relief.
Suppose 3 policy scheme available with L.I.C. and 5 policy
schemes are available with Birla life insurance, in how many
ways this can be done ?.
ANS Using ADDITION PRINCIPLE we have 3+ 5=8 choices
Note (1) that here first operation is to get policy from L.I.C.
which can be done by 3 ways .
And second operation means to get policy from Birla Life
Insurance .
(2) The meaning of words I operation and II operation
changed according to the problem asked

4. • Example 1 Suppose Rakesh decide to go Vapi and see movie
with his friends . He can go Vapi by 3 ways by car ,by auto or by
bus and suppose 5 different movies are running in cinema hall.
• In how many ways he can go with his friends to Vapi and see
movie ?
• Ans first operation (to go Vapi) can be done by 3 ways (car,
bus,auto)
• Second operation (to see a movie) can be done by 5 ways
(M1,M2,M3,M4,M5)
• Therefore he can go Vapi with his friends and see movie by
3x5=15 different ways

5. • EXAMPLE 2
• 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 ways
• After this unit place can be filled by 2 ways
• Total 3 digits no we can form =4x3x2= 24
Hundred
place
Tenth
place
Unit
place

6. • SECOND PART
• To 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

7. • PERMUTATION :
• 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 is
important . ABC ≠ ACB
A CB
A
A
A
A
A
B
B
B
B
B
C
CC
C
C

8. • 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

9. • 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 = n Pn
=
=
= n!

10. EXAMPLE :- How many 3 digits number can be formed using
digits 1,6,8,9,3,7 without repeating any digits ?
Remark : 3 digits numbers may be 697,737,. . . . .etc .Here order
of digits does matter
ANS : we have total n=6 objects (digits ) and out of these 6
objects we have to select r=3 objects (digits) and arrange to
form 3 digits number .
Hence the number of such 3 digits number = 6P3
=
= 120

11. • Three posts chairman vice chairman and secretary are to be
filled out of 10 suitable candidates . In how many different
ways these posts can be filled.
• Remark :Anyone out of these 10 candidates become chairman
similarly anyone of them can become vice chairman or
secretary
• Solution :
• First chairman can be selected by 10 ways
• After this vice chairman can be selected by 9 ways ( because
now only 9 candidates remains)
• After this secretary can be selected by 8 ways
• Therefore total number of different ways these posts can be
filled is =10x9x8
• =720

12. • 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

13. • 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

14. • COMBINATION
• 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)

15. 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
AB,BC,CA
PERMUTATIONS
AB,BA,BC,CB,AC,CA,

16. • 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
•

17. • EXAMPLE A Cricket team of eleven (11) players is to be
formed from 20 players consisting of 7 bowlers , 3 wicket
keepers and 10 batsmen.
In how many ways the team can be formed so that it contains
exactly 4 bowlers and 2 wicket keepers?
Solution :- 4 bowlers can be selected out of 7 by = 7C4 ways
2 wicket keepers can be selected out of 3 by= 3C2 ways
Remaining 6 batsman can be selected out of 10 by
= 10C5 ways
Hence total number of ways = 7C4 x 3C2 x 10C5

18. • EXAMPLE
• 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 x5C1
• 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