Combinatorics Through Guided Discovery, 2004a

185
207. In the power series ∑ ∞
j=0 q2ij ,the2ij has a different interpretation if you think
of it as (2i) · j or if you think of it as i · (2j).
208. Note that
1 − q 2
1 − q · 1 − q4
1 − q 2 · 1 − q6
1 − q 3 · 1 − q8
1 − q 4 = (1 − q6 )(1 − q 8 )
(1 − q)(1 − q 3 ) .
209. Note that q i + q 3i + q 5i + ···= q i (1 + q 2 + q 4 + ···).
210.a. We want to calculate the number of partitions whose Young diagrams fit into
a two by two square. These partitions have at most two parts and the parts
have size at most two. Thus they are partitions of 1, 2, 3, or 4. However not
all partitions of 3 or 4 have diagrams that fit into a two by two square. Try
writing down the relevant diagrams.
210.b. They are the generating function for the number of partitions whose Young
diagram fits into a rectangle n−1 units wide and 1 unit deep or into a rectangle
1 unit wide and n − 1 units deep respectively.
210.c. How can you get a diagram of a partition counted by partition is counted by
from one whose partition is counted by [m+n
m ] q ?
[ m+n
n
] q
210.e.iii. Think about geometric operations on Young Diagrams
210.f. How would you use the Pascal recurrence to prove the corresponding result
for binomial coefficients?
210.g. For finding a bijection, think about lattice paths.
210.h. If you could prove [ m+n
n
] is a polynomial function of q, what would that tell
q
you about how to compute the limit as q approaches −1?
Additional Hint: Try computing a table of values of [ m+n
n
] with q = −1 by
q
using the recurrence relation. Make a pretty big table so you can see what is
happening.
211.c. You may run into a product of the form ∑ ∞
i=0 ai x i ∑ ∞
j=0 b j x j . Note that in the
product, the coefficient of x k is ∑ k
i=0 ai b k−i = ∑ k
i=0 ai .
b i
214. Our recurrence becomes a n = a n−1 + a n−2 .
217.
gives us
5x +1 cx +5c + dx − 3d
=
(x − 3)(x − 5) (x − 3)(x − 5)
5x = cx + dx