a) 1, 2 ∈ A1 and 3, 4 ∈ A2 and 5, 6, 7 ∈ A3
b) 1, 2 ∈ A1 and 3, 4 ∈ A2 and 5,6 ∈ A3 and |A1| = 3, where |A1| is the cardinality of A1
c) 1,2 ∈ A1 and 3,4 ∈ A2 and 5,6 ∈ A3
How would I go about doing this? Thanks !:)
b) 1, 2 ∈ A1 and 3, 4 ∈ A2 and 5,6 ∈ A3 and |A1| = 3, where |A1| is the cardinality of A1
c) 1,2 ∈ A1 and 3,4 ∈ A2 and 5,6 ∈ A3
How would I go about doing this? Thanks !:)
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In (a) there are only three ways, corresponding to the partition A_1 = {1,2,8}, A_2 = {3,4}, A_3 = {5,6,7}, the partition A_1 = {1,2}, A_2 = {3,4,8}, A_3 = {5,6,7}, and the partition A_1 = {1,2}, A_2 = {3,4}, A_3 = {5,6,7,8}. The point is that the given requirements almost completely determine the sets A_1 and A_2 and A_3; we just need to choose which set 8 goes into, and there are three different choices we can make in doing that.
In (b) there are 4 ways. The reason for this is that to make a partition subject to these constraints, we must add exactly one more element to A_1 besides 1 and 2. Since 3,4,5,6 are accounted for by the constraints, this one more element must be either 7 or 8. There are two ways to make that choice. Having done that, there are then two different places we can put the other number (ie, if we put 7 in A_1, there are two places we can put 8, in A_2 or in A_3; and this sentence remains true if you interchange 7 and 8. So, four partitions in all, corresponding to the choices "7 in A_1, 8 in A_2", "7 in A_1, 8 in A_3", "8 in A_1, 7 in A_2", and "8 in A_1, 7 in A_3."
In (c) there are 9 different partitions. The point is that 1, 2, 3, 4, 5, 6 are accounted for already so a partition that meets the constraints is entirely determined by where we put 7 and 8. There are 3 places we can put 7, and 3 places we can put 8, and these choices can be made entirely independently of one another (ie, we could put 7 in A_1 and 8 in A_1, or 7 in A_1 and 8 in A_2, or 7 in A_1 and 8 in A_3, or 7 in A_2 and 8 in A_1, or 7 in A_2 and 8 in A_2, or 7 in A_2 and 8 in A_3, or 7 in A_3 and 8 in A_1, or 7 in A_3 and 8 in A_2, or 7 in A_3 and 8 in A_3. These are nine possible choices, and they all lead to different partitions).
As for "how you go about doing this," really it's just counting, and how you organize the counting is up to you. If you're in a class where you've learned about various theorems that count things, maybe you can organize the count using those theorems. But the basic idea, however you count, is to notice how the structure of the problem (making a partition of A, subject to constraints) means that in each situation you are just counting a certain number of things. In (a) you're counting the number of places you can put 8, in (b) you're essentialy counting the number of pairs (a,b) where a is what you put in A_1 and b is either 2 or 3 and determines where you put the thing that you didn't put in A_1, and in (c) you're essentially counting the number of pairs (a,b), where a runs from 1 to 3 and b runs from 1 to 3.
In (b) there are 4 ways. The reason for this is that to make a partition subject to these constraints, we must add exactly one more element to A_1 besides 1 and 2. Since 3,4,5,6 are accounted for by the constraints, this one more element must be either 7 or 8. There are two ways to make that choice. Having done that, there are then two different places we can put the other number (ie, if we put 7 in A_1, there are two places we can put 8, in A_2 or in A_3; and this sentence remains true if you interchange 7 and 8. So, four partitions in all, corresponding to the choices "7 in A_1, 8 in A_2", "7 in A_1, 8 in A_3", "8 in A_1, 7 in A_2", and "8 in A_1, 7 in A_3."
In (c) there are 9 different partitions. The point is that 1, 2, 3, 4, 5, 6 are accounted for already so a partition that meets the constraints is entirely determined by where we put 7 and 8. There are 3 places we can put 7, and 3 places we can put 8, and these choices can be made entirely independently of one another (ie, we could put 7 in A_1 and 8 in A_1, or 7 in A_1 and 8 in A_2, or 7 in A_1 and 8 in A_3, or 7 in A_2 and 8 in A_1, or 7 in A_2 and 8 in A_2, or 7 in A_2 and 8 in A_3, or 7 in A_3 and 8 in A_1, or 7 in A_3 and 8 in A_2, or 7 in A_3 and 8 in A_3. These are nine possible choices, and they all lead to different partitions).
As for "how you go about doing this," really it's just counting, and how you organize the counting is up to you. If you're in a class where you've learned about various theorems that count things, maybe you can organize the count using those theorems. But the basic idea, however you count, is to notice how the structure of the problem (making a partition of A, subject to constraints) means that in each situation you are just counting a certain number of things. In (a) you're counting the number of places you can put 8, in (b) you're essentialy counting the number of pairs (a,b) where a is what you put in A_1 and b is either 2 or 3 and determines where you put the thing that you didn't put in A_1, and in (c) you're essentially counting the number of pairs (a,b), where a runs from 1 to 3 and b runs from 1 to 3.