Combinatorics Through Guided Discovery, 2004a

182 D. Hints to Selected Problems
171.c. Here the harder part requires that, after removal, you consider a range of
possible numbers being partitioned and that you give an upper bound on the
part size. However it lets you describe the number of parts exactly.
171.d. One of the two sets of partitions of smaller numbers from the previous part
is more amenable to finding a recurrence than the other. The resulting recurrence
does not have just two terms though.
171.h. If there is a sum equal to zero, there may very well be a partition of zero.
172. How does the number of compositions of k into n distinct parts compare to
the number of compositions of k into n parts (not necessarily distinct)? What
do compositions have to do with partitions?
173. While you could simply display partitions of 7 into three parts and partitions
of 10 into three parts, we hope you won’t. Perhaps you could write down the
partitions of 4 into two parts and the partitions of 5 into two distinct parts
and look for a natural bijection between them. So the hope is that you will
discover a bijection from the set of partitions of 7 into three parts and the
partitions of 10 into three distinct parts. It could help to draw the Young
diagrams of partitions of 4 into two parts and the partitions of 5 into two
distinct parts.
174. In the case k =4and n =2,wehavem =5. In the case k =7and n =3,we
have m =10.
175. What can you do to a Young diagram for a partition of k into n distinct parts
to get a Young diagram of a partition of k − n into some number of distinct
parts?
176. For any partition of k into parts λ 1 , λ 2 , etc. we can get a partition of k into
odd parts by factoring the highest power of two that we can from each λ i ,
writing λ i = γ i · 2 k i .Whyisγ i odd? Now partition k into 2 k 1
parts of size γ 1 ,
2 k 2
parts of size γ 2 , etc. and you have a partition of k into odd parts.
177. Suppose we have a partition of k into distinct parts. If the smallest part, say m,
is smaller than the number of parts, we may add one to each of the m largest
parts and delete the smallest part, and we have changed the parity of the
number of parts, but we still have distinct parts. On the other hand, suppose
the smallest part, again say m, is larger than or equal to the number of parts.
Then we can subtract 1 from each part larger than m, and add a part equal
to the number of parts larger than m. This changes the parity of the number
of parts, but if the second smallest part is m +1, the resulting partition does
not have distinct parts. Thus this method does not work. Further, if it did
always work, the case k 3j2 +j
2
would be covered also. However you can
modify this method by comparing m not to the total number of parts, but to
the number of rows at the top of the Young diagram that differ by exactly one