Question 1208798
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Construct two example sets
A = {a, b, c, d}
B = {e, f}


A x B = <a href="https://mathworld.wolfram.com/CartesianProduct.html">cartesian product</a>
A x B = set of ordered pairs of form (p,q) where p is from set A and q is from set B
A x B = {
(a,e), (a,f),
(b,e), (b,f),
(c,e), (c,f),
(d,e), (d,f)
}


If you want you can flatten the set into one single line like this
A x B = { (a,e), (a,f), (b,e), (b,f), (c,e), (c,f), (d,e), (d,f) }
But it might be preferable to have it broken up into rows.


4 items in set A, 2 items in set B, n = 4*2 = 8 ordered pairs in AxB.
There are 2^n = 2^8 = 256 possible subsets of AxB (see <a href="https://en.wikipedia.org/wiki/Power_set">Power Set</a>)
This includes the set AxB itself and the empty set.
Refer to Pascal's Triangle. Look at the row that starts with "1,8,28...". Each value in Pascal's Triangle can be computed using the nCr combination formula.
All of those values in that row add to 256. 


The first three items of this row correspond to:<ul><li>1 empty set</li><li>8 subsets that are <a href="https://mathworld.wolfram.com/SingletonSet.html">singletons</a>. These are the subsets that have exactly one ordered pair.</li><li>28 subsets that have two ordered pairs</li></ul>Add up those first three values to get 1+8+28 = 37
There are 37 subsets with at most 2 ordered pairs.
Then 256-37 = <font color=red>219</font> is the amount of subsets that have between 3 ordered pairs and 8 ordered pairs.


Answer: <font color=red>Choice A</font>
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