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Permutation and
Combination
AN INTRODUCTION
Groups And Arrangements
 Suppose we have three letter a,b,c and we are
asked to form groups consisting of two letters
each.
 The groups are ab,ac,bc.
 Now the letters of each one of these groups can be
arranged in two different ways and we get the
following arrangements: ab,ba;ac,ca;bc,cb.
 In forming the groups we ignore the order in
which the letters occur. But in making
arrangements we also take into account the order
in which the letters are to be placed.
Permutation And Combination
 Permutation: Each of the different
arrangements which can be made by taking some
or all the numbers or things at a time is called a
permutation.
 Combination: Each of the groups or selections
which can be made by taking some or all of a
number of things at a time is called a
combination.
 That means “permutation” is related with
arrangements and “combination” is related with
groups.
Permutation And Combination
Permutation And Combination
 We use the symbol nPr or P(n,r) to denote the
number of “permutations” of n different things
taken r at a time.
 And we use the symbol nCr or C(n,r) to denote the
number of combinations of n different things taken
r at a time. rW, n N, 0≤r≤n.
 Now nPr= n/ (n-r) and
 nCr = n/ (  r (n-r) )
Some facts about Permutation
 nPr=n(n-1)(n-2)(n-3)……upto r factors.
 nPr=n(n-1)(n-2)(n-3)……..(n-r+1)
 nPn= n and nP1=n.
Examples
 A lady wants to select one cotton saree and one
polyester saree from a shop. If there are 10 cotton
varieties and 12 polyester varieties, in how many ways
can she choose the two sarees?
 There are 12 buses running between Una and Amb. In
how many ways can a man go from Una to Amb and
return by a different bus?
 The license plates for vehicles registered in Delhi consist
of 3 letters (of English alphabet) followed by 1,2,3, or 4
digits. The letter on the extreme left has to be ‘D’. For the
1-digit number plates, the number 0 is not allowed. For
others, the digits and the letters, of course, can repeat,
but the numbers should be significant. Determine the
possible number of license plates.
Solution
 A sequence of three letters beginning with D can be
chosen in 26 x 26=676 ways as explained below:
First Letter
Second Letter
Third Letter
D
26
ways
26
ways
1
26
26
676
Solution
 Corresponding to each sequence of four digits, there
are 9999 license plates possible as any number from
1 to 9999 can be allotted.
 Hence, the possible number of license plates
=676 x 9999=676(10000-1)
=6760000-676=6759324
Fundamental Principle of Counting
(Association)
 If one operation can be performed in m different
ways and if corresponding to each way of
performing this operation, there are n different
ways of performing second operation, the two
operations together can be performed in m x n
ways.
 Generalization: if there are p different ways to
performing a third operation corresponding to each
of m x n ways of performing the first two
operations, then the three operations together can
be performed in m x n x p ways, and so on.
Exercise 1.1
 How many numbers are there between 100 and




1000 in which all the digits are distinct?
How many numbers are there between 100 and
1000 such that every digit is either 2 or 5?
How many numbers are there between 100 and
1000 such that the digit in the unit’s place is 7.
How many numbers are there between 100 and
1000 such that at-least one of the digits is 8.
How many numbers are there between 100 and
1000 such that exactly one of the digits is 5.
Exercise1.2
 Find the value of
20P
3
Answer:
6840
 Find the value of n if n P4 : n-1 P3 =9:1
Answer:
n=9
 Find the value of n if P(n,4)=20P(n,2)
Answer:
n=7
Combinations
 To find the number of combinations of n different





things taken r at a time can be represented by nCr
and nCr= n/rn-r, where 0≤r≤n
Some facts about combinations:
nC =1
n
nC =1
0
nC = nC
r
n-r
nC = nC =n
n-1
1
Practical Problems on Combinations
 To find the number of ways in which m+n
things can be divided into two groups
containing m and n things respectively.
 C(m+n,n)=m+n/mn
ways
 To find the number of ways in which a
selection can be made out of p+q+r things of
which p are alike of one kind, q are alike of
another kind and remaining r are different.
 =(p+1)(q+1)2r -1
ways
Exercise 1.3
In how many ways can a Football Eleven be
selected out of 15 players?

1.
2.
In how many of them a particular player is included.
In haw many of them he is excluded?
In first situation we have to include a particular player, so we
have 14 players and we have to choose 10 players out of
those remaining 14 players so the combination is 14C10
In second situation we have to excluded a particular player, so
we have 14 remaining players and we have to choose 11
players out of those remaining 14 players so the
combination is 14C11
Exercise 1.3
A committee of 6 members is to be formed from 6
boys and 4 girls. In how many ways can this be
done if the committee contains

2 girls ?
2.
atleast 2 girls ?
In first case exactly 2 girls are there in the committee. So we
have to choose in total 6 members. 2 girls out of 4 girls
present. And remaining 4 boys out of 6 boys present. So by
the principle of association the required number of ways
are 6C4 X 4C2 =90
In second case atleast 2 girls means there can be 2,3 or 4 girls in
the committee. So the required number of ways are 6C4 X
4C + 6C X 4C + 6C X 4C =185
2
3
3
2
4
1.