Question 202395
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The answer is going to depend on whether you want to count all of the subsets or just the proper subsets.


For your first problem, the simplest and easiest way to do it is to look at the 5th row of Pascal's Triangle, namely 1 5 10 10 5 1, which tells us that:


There is 1 way to select 0 things from 5 things,


There are 5 ways to select 1 thing from 5 things,


There are 10 ways to select 2 things from 5 things,


There are 10 ways to select 3 things from 5 things,


There are 5 ways to select 4 things from 5 things, and


There is 1 way to select 5 things from 5 things -- all where order doesn't matter.  Order doesn't matter because sets are really just a bag of things rather than an ordered list; that is to say that {1, 2, 3}, {3, 2, 1}, and {2, 3, 1} are all equal.


So there are 1 + 5 + 10 + 10 + 5 + 1 = 32 subsets.  Now if you only want the proper subsets, you would eliminate one of these because the total includes the set itself as a subset.  The set itself is a subset, just not a proper subset.  So, the number of proper subsets is 31.


You would solve your problem for set B in the same manner, except that you would use the 9th row of Pascal's triangle, the coefficients of the binomial expansion of *[tex \Large (a + b)^9], or calculate the combinations *[tex \Large _rC_9] for *[tex \Large r = 0] through *[tex \Large r = 9].


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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