Answer:
How many solution does ++=11
x
+
y
+
z
=
11
have where ,,
x
,
y
,
z
are non-negative integers. In light of the restrictions, its clear that ,,∈{0,1,2,..11}
x
,
y
,
z
∈
{
0
,
1
,
2
,
.
.11
}
. So, at face value I would assign a value for
x
and determine the different combinations that
y
and
z
can hold. For example,
For =0
x
=
0
, we have +=11
y
+
z
=
11
. With writing them out I found that there are 12
12
different assigned combinations for
y
and
z
that satisfy the equation. For =1
x
=
1
, I got 11
11
. Consequently, the pattern becomes clear whereby each one takes a value less by one. Hence, the number of solutions is 1+2+3+4+5+6+7..+12=78
1
+
2
+
3
+
4
+
5
+
6
+
7..
+
12
=
78
. I was wondering if there is an easier method perhaps with combinations equation (,)
C
(
a
,
b
)
..?
Step-by-step explanation:
