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Precalculus Mathematics Homework Help
Conditional distribution
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[QUOTE="nuuskur, post: 4990502, member: 519618"] If I simplify - assume that there is a child that receives only one present, then by the condition: Per every 2 child, they must have a common present. This means that everybody has to have the same gift as the child with one gift. looks something: [tex]C_1 = \{x_1\}\\C_2 = \{x_1, x_2\}\\C_3 = \{x_1, x_2, x_3\}[/tex] I propose that: if the maximum amount of people are attending, then there is one person who receives one present there can be Only one person who receives one present. If there are 2 people who receive one present, then they do not have a common present, because no two sets can be identical and that also violates the next condition. IF the smallest set of gifts has 2 presents, then there has to be a present x[SUB]i[/SUB] that is in every set of presents, but then we can have another person at the party who receives the gift x[SUB]i[/SUB], therefore if the maximum amount of people are attending, the smallest set of gifts contains one present. For 2 gifts, if x[SUB]i[/SUB] is present, there are also n-1 possibilities for the second gift. For 3 gifts, if x[SUB]i[/SUB] is present, there are [itex]\binom{n-1}{2}[/itex] possibilities to combine the sets of gifts that contain x[SUB]i[/SUB] . . [I]Careless mistake - I had written [itex]\sum_{k=0}^n \binom{n-1}{k}[/itex], but after k = n we get error :D[/I] There would be [itex]\sum_{k=0}^{n-1}\binom{n-1}{k} = 2^{n-1}[/itex] possible ways - the maximum amount of people attending. Now I see an easier way - we must consider the subsets that contain x[SUB]i[/SUB] - that leaves n-1 types to choose from. The number of subsets of n-1 types is 2[SUP]n-1[/SUP] [/QUOTE]
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Conditional distribution
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