PERMUTATIONS & COMBINATIONS

Fundamental principle of Counting:
We have two principles of counting the “Sum rule” and the “Product rule”.
Event A can happen in = m ways.
Event B can happen in = n ways.
Then, the chances of either (A or B) happening are given by ⎯⎯⎯⎯⎯→ can occur in m + n ways (Sum Rule)
The chances of both (A and B) happening simultaneously are given by ⎯⎯⎯⎯⎯→ can occur in m × n ways (Product
Rule)
Concept of nPr and nCr:
r
nP → Number of arrangements of n different objects taken ‘r’ at a time
nCr → Number of selections of n different objects taken ‘r‘ at a time.
Formulae of Permutation:
1. Permutations of n different things taken 'r' at a time is denoted by nPr and is given by
nPr =
(n r)!
n!
−
2. The total number of arrangements of n things taken ‘r’ at a time, in which a particular thing always
occurs = n–1Pr–1.
E.g. The number of ways in a basketball game in which 5 out of 8 selected players can play at different
positions, such that the captain always plays at the centre position = 8 – 1 P5 – 1 = 7P4 = 210.
3. The total number of permutations of n different things taken ‘r’ at a time in which a particular thing never
occurs = n–1Pr.
E.g. The number of ways in which we can form a 4 letter word from the letters of the word COMBINE
such that the word never contains B = 7–1P4 = 6P4 = 30
4. The number of arrangements when things are not all different such as arrangement of n things, when p
of them are of one kind, q of another kind, r is still of another kind and so on, the total number of
permutations is given by
(p! q! r !.......)
n! .
E.g. The total arrangements of the letters of the word “M A T H E M A T I C S” in which M, A and T are
repeated twice respectively =
2!2!2!
11!
5. The number of permutations of n different things taking ‘r’ at a time when each thing may be repeated
any number of times in any permutations is given by (n × n × n × n × n……………..r times) i.e. nr ways.

E.g. The total numbers of ways in which 7 balls can be distributed amongst 9 persons (when any man
can get any number of balls) = 97 ways.
Circular Permutations:
In linear permutation, we fill first place by n ways and next in (n – 1) ways and so on, but in circular
arrangement we don’t have any first place. So fix any object as a first place and arrange the rest (n – 1)
objects around it. Hence, we have to arrange 1 less than the total number of things.
i. Number of circular permutations of n things all taken at a time = (n – 1)!
ii. Number of circular permutations of n different things taking ‘r’ at a time =
r
Pr
n
.
iii. If there is no difference between clockwise and anticlockwise arrangements, the total number of circular
permutations of n things taking all at a time is
2
(n −1)!
& the total number of circular permutations n
when taking ‘r’ at a time all will be
2r
Pr
n
.
Formulae of Combination:
1. Number of combinations of n dissimilar things taken 'r' at a time is denoted by nCr & is given by
nCr =
(n r)!r!
n!
−
2. Number of combinations of n different things taken ‘r’ at a time in which ‘p’ particular things will always
occur is n–pCr–p
E.g. The number of ways a basketball team of 5 players chosen from 8 players, so that the captain be
included in the team = 8 – 1 C5 – 1 = 7C4 = 35
3. Number of combinations of n dissimilar things taken 'r' at a time in which 'p' particular things will never
occur is n–pCr
E.g. The number of ways a basketball team of 5 players chosen out of 10 players, such that the player
named Saurav should not be included in the team = 10 – 1 C5 = 9C5 = 126.
4. The number of ways in which (m + n) things can be divided into two groups containing m & n things
respectively (m + n)Cn =
m! n!
(m + n)!
= (m + n)Cm

5. If 2m things are to be divided into two groups, each containing m things, the number of ways =
[2(m! ) ]
(2m)!
2 .
6. The number of ways to divide n things into different groups, one containing p things, another q things &
so on is equal to
p!.q!.r!......
(p + q + r + .....)!
Where {n = p + q + r +…}
Distribution of Identical Objects:
The total number of ways of dividing n identical items among r persons, each of whom can receive 0, 1, 2, or
more items (≤ n) is n + r – 1Cr – 1.
OR
The total number of ways of dividing n identical objects into r groups, if blank groups are allowed, is
n + r – 1Cr – 1.
Example:
How many non – negative integral solutions are possible for the given equation?
x + y + z = 16
Hint: Here, if we look out to the problem we will find 16 objects have to be distributed among 3 different
persons (i.e. x, y, z).
Hence n = 16, r = 3 and total number of non – negative solutions = 3 1
16 3 1C −
+ − = 2
18C = 153.
Some important Results:
i. Number of lines with n points = nC2.
Q For making a line exactly two points are required. So the number of ways in which we can choose
two points out of n point is nC2.
Combination is used here because a line from A to B is the same as from B to A. So AB & BA are the
same.
(i) n lines can intersect at a maximum of nC2 points.
(ii) Number of triangles with n points = nC3.
(iii) Number of diagonals in n sided polygon = nC2 – n
ii. The number of ways in which mn different items can be divided equally into m groups, each containing
n objects and the order of the groups is not important, is
m!
1
(n! )
(mn)!
m ⎟
⎟
⎠