CSE 21: Solutions  Problem Set 4
1.
These problems are variations of
Bars and Stars
. Let
x
1
+
x
2
+
x
3
+
x
4
=
n
be the equation.
(a) This is of the standard form: there are
n
stars, and we need 3 bars. The answer is
(
n
+3
3
)
.
(b) If all four variable are positive, transform the equation to (
x
1

1)+(
x
2

1)+(
x
3

1)+(
x
4

1) =
n

4. Change variables:
x
i

1
→
y
i
. Then
y
1
+
y
2
+
y
3
+
y
4
=
n

4. There is a bijection between
positive solutions to
x
i
’s and nonnegative solutions to
y
i
’s.
Thus the number of solutions (to
both equatiosn) is
(
n

4+3
3
)
. The probability is
(
n

1
3
)
/
(
n
+3
3
)
.
(c) Transform the equation to (
x
1
+2)+(
x
2
+2)+(
x
3
+2)+(
x
4
+2) =
n
+8. The answer is
(
n
+8+3
3
)
.
(d) Transform the equation to (8

x
1
) + (8

x
2
) + (8

x
3
) + (8

x
4
) = 32

n
.
The answer is
(
32

n
+3
3
)
.
2.
(a) Let
m
be the size of the multisets, and
n
be the alphabet size. As shown in class, we can either
use
Stars and Bars
, or transform it into counting subsets. The answer is
(
m
+
n

1
n

1
)
.
(b) Let the functions be from [
m
] to [
n
]. It’s easy to see that the collection of function values is a
multiset of size
m
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 Spring '07
 Graham
 stage, Monotonic function, Convex function

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