Download 26) Let hn denote the number of such colorings, where

26) Let h n denote the number of such colorings, where h 0 = 1. Since this problem is a
permutation problem (ordering matters here), we use the exponential generating function.
We want to color a 1 × n chessboard using the colors red, blue, green and orange. Since
even numbers of squares are to be colored red and also even numbers of squares are to be
colored green, then their exponential generating function is
2
4
1+ x + x +⋯
2!
4!
Since there is no restriction how we suppose to use the colors blue and orange, then their
generating function is
2
3
1+x+ x + x +⋯
2!
3!
Then the generating function for this problem is
2
2
4
2
3
1+ x + x +⋯
1+x+ x + x +⋯
2!
4!
2!
3!
2
= 1 e x + e −x  e x  2
2
2
= 1 e 2x + 1
2
1
= e 4x + 2e 2x + 1
4
2
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g e =
∞
∞
= 1
4
1+∑
= 1
4
n
n+1
1+∑ 4 +2
n!
n=0
∞
4nxn + 2 ∑ 2nxn
n!
n!
n=0
xn
n=0
∞
n
n+1
= 1 +∑ 4 +2
4n!
4
xn
n=0
Hence h n for n ≥ 1 is given by the coefficent of x n multiply by n!, i.e. h n =
4 n +2 n+1
4
.
27) Let h n be the number of ways creating an n digit numbers with the requirements: all
digits are odd, 1 and 3 each occur a nonzero, even number of times. This problem also a
permutation problem, hence we use the exponential generating function. The set of odd
digit numbers are 1, 3, 5, 7, 9. For each 1 and 3, we want them to appear nonzero, even
number of times. Hence the exponential generating function for each of this number is
x2 + x4 + ⋯
2!
4!
For the other numbers ( 5, 7 and 9 ) there is no restriction, hence the exponential generating
function for each of them is
2
1+x+ x +⋯
2!
Now the exponential generating function for this problem is
x2 + x4 + ⋯ 2 1 + x + x2 + ⋯
2!
4!
2!
2
2
4
=
1 + x + x + ⋯ − 1 e x  3
2!
4!
= 1 e x + e −x  − 1 e 3x
2
1
= e 4x + 1 e 2x − e 3x
2
2
g e =
∞
= 1
2
n
=
∑
n=0
Therefore h n =
4n
2
+
2n
2
∑
n=0
4nxn + 1
n!
2
∞
∑
n=0
4n + 2n − 3n
2
2
3
∞
2nxn − ∑ 3nxn
n!
n!
n=0
xn .
n!
− 3 n = 2 ⋅ 4 n−1 + 2 n−1 − 3 n .
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51) We want to solve reccurence relation
h n = 3h n−1 − 4n,
n ≥ 1
h0 = 2
Let gx = h 0 + h 1 x + h 2 x + ⋯ + h n x n + ⋯. Note that from the above we have
3h n−1 − h n = 4n. So now we want to create two series. In one series we want the coefficient
of x n to be 3h n−1 − h n and on the other series coefficient of x n is 4n. To create the first series
note that
2
3xgx = 3h 0 x + 3h 1 x 2 + 3h 2 x 3 + ⋯ + 3h n−1 x n + ⋯
Hence
3xgx − gx = −h 0 + 3h 0 − h 1 x + 3h 1 − h 2 x 2 + ⋯ + 3h n−1 − h n x n + ⋯
3x − 1gx = −2 + 3h 0 − h 1 x + 3h 1 − h 2 x 2 + ⋯ + 3h n−1 − h n x n + ⋯ (since h 0 = 2 )
By the reccurence relation, this series equal to
− 2 + 4x + 4 ⋅ 2x 2 + 4 ⋅ 3x 3 + ⋯ + 4nx n + ⋯ = −2 + 4x + 2x 2 + ⋯ + nx n + ⋯.
Note that
1
1−x
d
1
dx 1 − x
1
1 − x 2
x
1 − x 2
= 1 + x + x2 + x3 + ⋯
= 1 + 2x + 3x 2 + 4x 3 + ⋯
= 1 + 2x + 3x 2 + 4x 3 + ⋯
= x + 2x 2 + ⋯ + nx n + ⋯
Hence
4x + 4 ⋅ 2x 2 + 4 ⋅ 3x 3 + ⋯ + 4nx n + ⋯ = 4
Now since 3h n−1 − h n = 4n, then
x
−
1 x 2
3x − 1gx = −2 +
4x
1 − x 2
−21 − x 2 + 4x
1 − x 2
1
=−
2x 2 − 8x + 2
x − 1 2
3x − 1gx =
Now
2x 2 − 8x + 2
1 − 3x1 − x 2
2
1
=
− 1 +
2
x
3x
−
1
−1
x − 1
gx =
∞
∞
∞
n=0
n=0
= 2 ∑n + 1x n + ∑ x n − ∑ 3 n x n
n=0
∞
∑2n + 1 + 1 − 3 n x n
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=
n=0
∞
=
∑2n + 3 − 3 n x n .
n=0
Therefore
h n = 2n + 3 − 3 n .