MATH 460 2000/2001
Assignment 1
Due Oct. 10, 2001
1. Find the coefficient of x in the expansion of
8
(a) x −
(b) x x
10
(
1)
3
Solution
+
2
12
+1
:
(a) By the Binomial Theorem, the coefficient of x in x −
is
.
−
(b) The number of x s used to make up x must be even. We can have or x s. That is,
8
3
10
(
10
1)
2
8
4
Therefore the coefficient is
1)
0
x
12
2
(
2
4
+
12
12
+
4
12
10
2
1
x
3
2
x
= 45
2
3
2
10
2
= 1155
1
2. Find the value of a in the following expansion:
50
− x
7
6
12
x − x
2
7
a
=
+2
0
+
ax ax
+
1
2
2
+
··· a x
+
50
50
+
···
(Hint: Use partial fractions.)
Solution :
6
− x
x − x
7
12
2
7
2
=
+2
1
=
50
= 2
·
2
50
+
3
2
·
3
50
2
51
= 2
2
3
+
2
2
=
Therefore a
− x
1
− x
∞
2
k
51
2
+
+
3
2
− x
3
2 1
3
1
− x
∞
3
2
k xk + 3
2
k
2
=0
=0
1
2
1 2 3
1
1
2
=
2
x
1
+
x
2
The generating function for each xi is
1+
10
10
+
xk
100
1
0
and
2
10
a ≤ a ≤ ··· ≤ a
consisting of integers from the set { , , , ..., }.
a − , x a − a , ..., x a − a , x
Solution : Let x
xi ≥ , i , , ...,
=
k
.
3. (a) Find the number of nondecreasing sequences of length
1
3
=
+
x x
1
9
= 1 2
··· x
+
10
2
11
+
=
= 100
11
= 100
···
11
−
− x.
1
1
1 = 99
− a . We have
10
Therefore the solution is given by the coefficient of x in −x 11 . By the Generalized Binomial
Theorem,
∞
1
99
(1
1
−x
(1
=
11
)
k
10
k xk
10 +
10
=0
.
The solution is
(b) Find the number of increasing sequences of length
109
)
= 42 634 215 112 710
10
a < a < ··· < a
consisting of integers from the set { , , , ..., }.
1
2
1 2 3
10
100
: Since every selection of numbers
} form a unique increasing
from the set { , , , ...,
.
sequences of length , the solution is
(c) Find the number of ways to select ten distinct letters from the alphabet {A, B, C, ..., Z } if no two
consecutive letters can be selected.
Solution : Let the ten letters selected be a , a , ..., a
arranged in the alphabet order. Let x be
the number of letters before a , x be the number of letters between a and a , ..., x be the
number of letters between a and a and x be the number of letters after a . We have x ≥ ,
. The generating function is
x ≥ and xi ≥ for i , , ..., . Also x x · · · x
Solution
10
1 2 3
100
10
10
1
1
11
1
10
1
1
10
= 2 3
(1 +
2
2
9
0
100
= 17 310 309 456 440
11
10
2
)
+
2
10
10
1 +
x ··· · x x
+
2
2 +
···
+
9
+
x ·
9
=
1
0
11 = 16
=
(1
x ···
(1 +
x
+
9
−x
11
)
11
)
.
The solution is given by the coefficient of x . By the Generalized Binomial Theorem,
16
x
(1
9
−x
17
x
=
11
)
∞
9
k
10 +
10
=0
kxk
.
Let k , we have the solution
(d) There are r red (identical) cars and b black (identical) cars parked in a row. How many ways can
this happen such that no two red cars are next to each other?
Solution : Let C be the first red car, C the second red car, ..., Cr the r th red car in the order
that they are parked. Let x be the number of black cars before C , x the number of black
cars between C and C , ..., xr the number of black cars between Cr− and Cr and xr the
number of black cars after Cr . We have x ≥ , xr ≥ and xi ≥ for i , , ..., r. Also
b. The generating function is
x x · · · xr
= 7
10
= 19 448
1
2
1
1
1
1
0
1
1
+
2
+
2
2
+
+1
=
(1 +
x ··· · x x
2
+
2
+
)
+
0
+1
· · · r−
1
+1
1
= 2 3
xr− ·
x ··· r
xr − .
−x r
1
=
(1 +
+
)
+1
1
=
(1
Since
xr−
−x r
1
(1
)
∞ r k
k
r x
r−
= x
1
+1
+1
)
k
+
=0
Letting k b − r , we have the coefficient of xb to be b r , which is the solution.
4. In how many ways can n letters be selected from n A’s, n B’s, and n C ’s?
Solution : The generating function is
=
+1
+1
3
1+
x x
+
2
+
2
··· x n
+
2
3
=
=
2
−x n
−x
− xn
2
1
3
+1
3
2
+1
=
1
− x
3
2
n
xn
−x
4
+3
(1
2
1
=
1
1
2
+1
)
x
+3
−x n
−x
−x n
2
+1
(1
+2
3
3
)
6
+3
3
4
n
+2
−x
n
6
+3
∞ k
k
=0
+2
2
xk
The solution is given by the coefficient of x n inthe generating function. Let k
n n .
have the coefficient of x n to be n − · n
3
3
3
+2
+1
3
2
2
= 3
2
+3
= 3
n and n − we
1
+1
k
5. (a) Show that k k
for all positive integers k.
Proof: Straightforward.
(b) Use part (a) to find a formula for
2
=
+2
2
2
2
+2
1
Solution
2
+3
··· n .
+
2
+
: By part (a),
n
k
k
n
2
=
k
=1
(
(1)
(
6
2
+ 1) (2
k
k
n
+2
n
k
n
+ 1)
(1)
.
2
∞ k xk
k
2
=2
2
=
(
x
2
) =
x
) =
2
Ax
∞ k xk −
2
k
∞ l
=2
l
2
2
+2
2
=0
xl
x
2
=
(1
−x
n
(We used this function in Problem 4 as well.) Let bn
k
bn, B x , is A x multiplied by the summing operator.
=
)
(
(
)
=
=
Letting k n − ,
=
=1
3
)
k then the generating function of
2
)
Bx
2
(2)
Ax
−x
(
1
x
)
x
=
(1
∞ k
2
k
becomes
(
+3
4
)
∞ k
xk =
k
=0
∞ n
) =
n
+1
+ 3
3
xk
+2
(2)
3
=2
.
So bn
(c) Find a formula for
2
−x
3
=0
Bx
xn
+1
3
3
1
Solution
2
=1
=
2
=0
Letting l k − , we have
=
k
3
∞ k
xk
(
n
+1
) =
Ax
(
k
=1
+2
if k < . We can rewrite it as
= 0
=
because the generating function of k is
(
2
+ 1)
nn
1
Ax
since k
k
+2
2
=
n
k
=1
nn
=
We have the second term in
3
+2
3
+3
+
··· n .
+
3
: We use the same approach as in parts (a) and (b) for the sum of squares. First, since
n
3
n −n
3
=
3
6
2
2
+
n
3
n
we have
n
k
k
n
n
n
−
·
n
n − n.
3
=
6
=
6
n
3
=
k
=1
=
3
2
k
2
k − k
2
+3
3
=1
n k
3
2
+3
3
6
6
2
+
n
+3
3
k
Let A (x) be the generation function of k , i.e.,
=1
k
2
n
k −
2
2
=1
k
k
(3)
=1
3
Ax
(
∞ k ∞ k
k
x
xk
k
k
∞ k
xk−
x
)
=
=
3
=0
3
=
3
k
x
=
3
3
=3
3
∞ l
=3
l
+3
xl l k −
(
3
=0
=
3)
x
3
=
(1
−x
4
)
Multiplying by the summing operator, we have
Bx
(
where B x is the generating function of bn
we have
(
)
Bx
(
)
x
3
) =
(1
n
k
=
x
3
=
−x
k
∞ k
k
∞ n
=0
n
and bn
=
4
.
(3)
xk
xk
+4
4
+1
+3
4
=3
n+1
+4
4
=0
=
3
∞ k
=
k. By the Generalized Binomial Theorem,
=1
5
)
xn
becomes
6
=
=
6
1
4
n
+1
+3
4
· n
+1
6. (a) Find a combinatorial proof of
(
k
2
+ 1)
k −
2
=1
+3
4
n n
2
n
· nn
1
6
(
n
2
k
k
=1
+ 1) (2
n
+ 1)
.
n m
m k
4
=
n
k
n−k
m−k
− ·n n
2
(
+ 1)
2
for any non-negative integers n ≥ m ≥ k.
Solution : In a group of n people, we will select m to form a committee. Among the committee
members, k of them will beexecutives. If we elect the committee members first, then select the
executives, there are mn mk possible outcomes. On the other hand,
if we elect the k executives
n−k n
first, then select the other m − k committee members, there are k m−k different ways this can
be done. Since the total outcomes must be the same, we have
n m
m k
(b) Evaluate
20
n
k
50
i
i
=0
Solution
=
n−k
m−k
− i.
−i
50
20
: By (a),
20
i
50
50
i
=0
20
− i
−i
20
=
i
=0
50
20
20
50
=
20
i
20
i
20
i
=0
50
=
20
2
20
.
(c) Evaluate
n n − nn −
k
k−
0
Solution
1
1
1
+
n n−
k−
2
2
+
2
···
k
−
+(
1)
n n − k .
k
0
:
k
i
(
−
i
=0
n n − i
i k−i
k
1)
=
n k
k i
i
k
n − i k
k i
i
−
(
i
1)
=0
=
(
1)
= 0
=0
7. Find the number of n-digit words generated from the alphabet { , , , , } in each of which the total
number of ’s and ’s is even.
Solution : For the total number of ’s and ’s to be even, there are either even number of ’s and
even number of ’s or there are odd number of ’s and odd number of ’s. Therefore the exponential
generating function is
0 1 2 3 4
0
1
0
1
1
0
0
ex e−x e x
2
+
3
2
+
ex − e−x e x
1
2
3
2
e x ex e−x ex − e−x e x e x e− x e x ex
3
=
+
4
2
+
3
=
The solution is
1
2
4
5
2
2
+2
2
=
2
+
2
n + 1).
(5
8. Find the number of n-digit words generated from the alphabet { , , } in each of which none of the
digits appears exactly three times.
0 1 2
5
Solution
: The exponential generating function is
ex − x
3
3
=
3!
e x− x e x
3
1
+
1
!
x ex −
2
3
=0
2
3
!
=0
1
2
2
(
=3
+
+
!
=0
216
1
12
+3
2
1
!
!
x
1
6
12
2
3
=0
=
2
∞ k xk
∞ k xk
−
x
k
k
k
k
∞ k xk ∞ k xk
−
k
k
k
k
∞
∞
n xn
n− xn
−
n
n
n n−
=0
=
3
2
=
1
3
3)!
x
12
+
6
∞ xk
k −
k
=0
12
1
!




n
n
3 −
n
3 −
n
3 −
0
n−3 n!
nn−−3
n
nn−−3
n
n−
1 2
2 (
3)!
1 2
!
2 (
3)!
1 2
!
2 (
3)!
+
+
1
12 (
1
12 (
n
n−
n
n−
3
!
6)!
!
6)!
6
−
6
9!
216
9
9
1
=6
(
6)!
!
3
x
216
The coefficient of xnn is





1
216
+6
=0
1
!
∞ xk
k − x
k
∞ xn
− x
n n−
1
3
9
≤n≤
≤n≤
2
5
≤ n, n n
= 9
= 9
216
9