On the right track, but means you can choose or again, either creating an overcount or making your blocks not a partition of . Think about why. (Hint: Think about what is supposed to represent).
I am understanding this a little better. Could someone let me know if I am on the right track with the following proof?
Give combinatorial proof: n 1 and k 1 then
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Condition on whether the element n is in a block by itself. If it is, then all such partitions can be constructed by first specifying a (k-1) partition of [n-1] and then adding the block {n}. There are S(n-1,k-1) such partitions.
We can continue the next case by putting n and n-1 into a block by themselves. We construct these partitions by first specifying a (k-1) partition of [n-2] and then adding the block {n,n-1}. There are such partitions. We can continue in this same manner until we reach a (k-1) partition of [n-n] by noting that S(n,k)=0 for all n<k. Then we use the sum principle for all terms.
Please note that in my text [n]={1,2,3,...,n}
Yes, thanks, I talked to my professor today. I am finally getting close to understanding these concepts. The major problem is that I restricted the second member to n-1 instead of any other member besides n. If we let the second member partitioned with n be any of the n-1 other elements, then the makes sense.