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Through the study of the restricted partition function PA(n) := # {(m1. ma) € zH4l all mj 2 0, 1[6 [ +maad =n} the number of partitions of using onlv the elem...

Question

Through the study of the restricted partition function PA(n) := # {(m1. ma) € zH4l all mj 2 0, 1[6 [ +maad =n} the number of partitions of using onlv the elements of a5 DA

through the study of the restricted partition function PA(n) := # {(m1. ma) € zH4l all mj 2 0, 1[6 [ +maad =n} the number of partitions of using onlv the elements of a5 DA



Answers

Consider the situation where $n$ items are to be partitioned into $k<n$ distinct subsets. The multinomial coefficients $\left(\begin{array}{c}n \\ n_{1} n_{2} \ldots n_{k}\end{array}\right)$ provide the number of distinct partitions where $n_{1}$ items are in group $1 . n_{2}$ are in group $2, \ldots ., n_{k}$ are in group $k$. Prove that the total number of distinct partitions equals $\left.k^{n} \text { . [Hint: Recall Exercise } 2.68 \text { (d). }\right]$

To proof that results, we can use the multi nominal expansion and this is a formula here we just a lot why one equals Y two. He calls dot dot, he calls y K. He calls what? So it becomes one plus one plus dot dot plus one. There are Kay wants into the power and he calls the song of Ben choose and one and two into NK and times one to a power 111 draw power to in a one to a power of in cake. So the left hand side is just okay to a power of end and the right hand side is some of and shoes and one and two. And um Kay know that all this is just one and this proved is that raised out of Libya.

Mhm for this problem we are told that if only regular partitions are allowed then we could not always partition and interval A to B in a way that automatically partitions sub intervals A and C inclusive and see and be inclusive. For sea between A and B. Were then asked why not? So keep in mind that if we are using a regular partition then we would have that delta X has to be b minus A over n. Which means that we will only have partition endpoints at some integer multiple. Like we could say I times some integer multiple of B minus a over end. If C cannot be expressed as I times B minus a over end, then inherently we'll be partition we won't be able to fully partition an interval that way for instance, let's say A equals zero, B equals one and equals two and C equals 3/4. Having that we can see that our sub intervals would be zero and one half And then 1/2 and one. We don't partition a sub interval, we can say that that does not ever equal. You know, it we will never be able to get or partition any sub interval of 0-3 over four


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