is given by
n
l
=
k
n
l
=
n
n
+
n
n

1
+
n
n

2
+
. . .
+
n
k
+ 1
+
n
k
.
Problem 6 (counting the number of increasing vectors)
If the first component
x
1
were to equal
n
, then there is no possible vector that satisfies the
inequality
x
1
< x
2
< x
3
< . . . < x
k
constraint.
If the first component
x
1
equals
n

1
then again there are no vectors that satisfy the constraint.
The first largest value that
the component
x
1
can take on and still result in a complete vector satisfying the inequality
constraints is when
x
1
=
n

k
+1 For that value of
x
1
, the other components are determined
and are given by
x
2
=
n

k
+ 2,
x
3
=
n

k
+ 3, up to the value for
x
k
where
x
k
=
n
.
This assignment provides
one
vector that satisfies the constraints. If
x
1
=
n

k
, then we
can construct an inequality satisfying vector
x
by assigning the
k

1 other components
x
2
,
x
3
, up to
x
k
by assigning the integers
n

k
+ 1
, n

k
+ 2
, . . . n

1
, n
to the
k

1
components. This can be done in
k
1
ways. Continuing if
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 Winter '08
 G.JOGESHBABU
 Counting, Probability, Natural number, Equals sign, X1, Total order, inequality satisfying vector, ﬁrst largest value

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