Moreover, if the sequence of
x
i
’s and
y
i
’s satisfies these inequalities, then the sequence of intersections
previously defined indicates a valid path. There are
(
5
k
)
ways of choosing the
k
tuple
(
x
2
, x
4
, . . . , x
2
k
)
and
(
4
k
)
ways of choosing the
k
tuple
(
x
1
, x
3
, . . . , x
2
k

1
)
subject to the required inequalities. Similarly,
there are
(
5
k
)
ways of choosing the
k
tuple
(
y
2
, y
4
, . . . , y
2
k
)
and because
y
2
k

1
could be equal to
y
2
k
+1
= 5
, there are
(
5
k
)
ways of choosing the
k
tuple
(
y
1
, y
3
, . . . , y
2
k

1
,
y
2
k
+1
)
. Therefore the total
number of possible paths is
4
∑
k
=0
(
5
k
)
3
(
4
k
)
= 1
3
·
1 + 5
3
·
4 + 10
3
·
6 + 10
3
·
4 + 5
3
·
1
= 1 + 500 + 6000 + 4000 + 125
= 10626
.
Diﬃculty:
Hard
NCTM Standard:
Geometry Standard for Grades 9–12:
use Cartesian coordinates and other
coordinate systems, such as navigational, polar, or spherical systems, to analyze geometric situations.
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 Winter '13
 Kramer
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