Part (c):
Consider the following manipulations of a binomial like sum
n
j
=
i
n
j
j
i
x
j

i
y
n

i

(
j

i
)
=
n
j
=
i
n
i
n

i
j

i
x
j

i
y
n

j
=
n
i
n
j
=
i
n

i
j

i
x
j

i
y
n

j
=
n
i
n

i
j
=0
n

i
j
x
j
y
n

(
j
+
i
)
=
n
i
n

i
j
=0
n

i
j
x
j
y
n

i

j
=
n
i
(
x
+
y
)
n

i
.
In summary we have shown that
n
j
=
i
n
j
j
i
x
j

i
y
n

j
=
n
i
(
x
+
y
)
n

i
for
i
≤
n
Now let
x
= 1 and
y
=

1 so that
x
+
y
= 0 and using these values in the above we have
n
j
=
i
n
j
j
i
(

1)
n

j
= 0
for
i
≤
n .
Problem 15 (the number of ordered vectors)
As stated in the problem we will let
H
k
(
n
) be the number of vectors with components
x
1
, x
2
,
· · ·
, x
k
for which each
x
i
is a positive integer such that 1
≤
x
i
≤
n
and the
x
i
are
ordered i.e.
x
1
≤
x
2
≤
x
3
≤
· · ·
≤
x
n
Part (a):
Now
H
1
(
n
) is the number of vectors with one component (with the restriction on
its value of 1
≤
x
1
≤
n
). Thus there are
n
choices for
x
1
so
H
1
(
n
) =
n
.
We can compute
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 Winter '08
 G.JOGESHBABU
 Addition, Vectors, Binomial, Probability, Natural number, following manipulations

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