Combinatorics Through Guided Discovery, 2004a

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181 162. How can you start with a partition of k and make it into a new partition of k +1that is guaranteed to have a part of size one, even if the original partition didn’t? 163. Draw a line through the top-left corner and bottom-right corner of the topleft box. 164. The largest part of a partition is the maximum number of boxes in a row of its Young diagram. What does the maximum number of boxes in a column tell us? 165. Draw all self conjugate partitions of integers less than or equal to 8. Draw all partitions of integers less than or equal to 8 into distinct odd parts (many of these will have just one part). Now try to see how to get from one set of drawings to the other in a consistent way. 166. Draw the partitions of six into even parts. Draw the partitions of six into parts used an even number of times. Look for a relationship between one set of diagrams and the other set of diagrams. If you have trouble, repeat the process using 8 or even 10 in place of 6. 167. Draw a partition of ten into four parts. Assume each square has area one. Then draw a rectangle of area 40 enclosing your diagram that touches the top of your diagram, the left side of your diagram and the bottom of your diagram. How does this rectangle give you a partition of 30 into four parts? 168.c. Consider two cases, m ′ > m and m ′ = m. 168.d. Consider two cases, n ′ > n and n ′ = n. 169. Suppose we take two repetitions of this complementation process. What rows and columns do we remove from the diagram? Additional Hint: To deal with an odd number of repetitions of the complementation process, think of it as an even number plus 1. Thus ask what kind of partition gives us the partition of one into one part after this complementation process. 170. How many compositions are there of k into n parts? What is the maximum number of compositions that could correspond to a given partition of k into n parts? 171.a. These two operations do rather different things to the number of parts, and you can describe exactly what only one of the operations does. Think about the Young diagram. 171.b. Think about the Young diagram. In only one of the two cases can you give an exact answer to the question.
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
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