DISCRETE MATHEMATICS
W W L CHEN
c
±
W W L Chen, 1992, 2008.
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Chapter 15
NUMBER OF SOLUTIONS
OF A LINEAR EQUATION
15.1. Introduction
Example 15.1.1.
Suppose that 5 new academic positions are to be awarded to 4 departments in the
university, with the restriction that no department is to be awarded more than 3 such positions, and
that the Mathematics Department is to be awarded at least 1. We would like to ﬁnd out in how
many ways this could be achieved. If we denote the departments by
M,P,C,E
, where
M
denotes
the Mathematics Department, and denote by
u
M
,u
P
,u
C
,u
E
the number of positions awarded to these
departments respectively. Then clearly we must have
u
M
+
u
P
+
u
C
+
u
E
= 5
.
(1)
Furthermore,
u
M
∈ {
1
,
2
,
3
}
and
u
P
,u
C
,u
E
∈ {
0
,
1
,
2
,
3
}
.
(2)
We therefore need to ﬁnd the number of solutions of the equation (1), subject to the restriction (2).
In general, we would like to ﬁnd the number of solutions of an equation of the type
u
1
+
...
+
u
k
=
n,
where
n,k
∈
N
are given, and where the variables
u
1
,...,u
k
are to assume integer values, subject to
certain given restrictions.
Chapter 15 : Number of Solutions of a Linear Equation
page 1 of 13
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View Full DocumentDiscrete Mathematics
c
±
W W L Chen, 1992, 2008
15.2. Case A – The Simplest Case
Suppose that we are interested in ﬁnding the number of solutions of an equation of the type
u
1
+
...
+
u
k
=
n,
(3)
where
n,k
∈
N
are given, and where the variables
u
1
,...,u
k
∈ {
0
,
1
,
2
,
3
,...
}
.
(4)
PROPOSITION 15A.
The number of solutions of the equation
(3)
, subject to the restriction
(4)
, is
given by the binomial coeﬃcient
±
n
+
k

1
n
²
.
Proof.
Consider a row of (
n
+
k

1) + signs as shown in the picture below:
+ + + + + + + + + + + + + + + +
...
+ + + + + + + + + + + + + + + +

{z
}
n
+
k

1
Let us choose
n
of these + signs and change them to 1’s. Clearly there are exactly (
k

1) + signs
remaining, the same number of + signs as in the equation (3). The new situation is shown in the picture
below:
1
...
1

{z
}
u
1
+
...
+ 1
...
1

{z
}
u
k

{z
}
k
(5)
For example, the picture
+1111 + +1 + 111 + 11 +
...
+ 1111111 + 1111 + 11
denotes the information
0 + 4 + 0 + 1 + 3 + 2 +
...
+ 7 + 4 + 2
(note that consecutive + signs indicate an empty block of 1’s in between, and the + sign at the lefthand
end indicates an empty block of 1’s at the lefthand end; similarly a + sign at the righthand end would
indicate an empty block of 1’s at the righthand end). It follows that our choice of the
n
1’s corresponds
to a solution of the equation (3), subject to the restriction (4). Conversely, any solution of the equation
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