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Let Set A = {1,2,3,4} & Set B = { a,b}
=> A x B contains 4*2 = 8 elements ( or ordered pairs),
Like A x B = { (1,a),(1,b),(2,a),(2,b),(3,a),(3,b),(4,a)(4,b) }
So, A x B contains 8 ordered pairs.
Now we need to find the subsets of A x B , in which at least 3 elements ie 3 ordered pairs should be there. That means it can have 8 ordered pairs or 7 ordered pairs or 6pairs or 5pairs or 4pairs or 3 pairs. It can not go below 3 pairs as question is we need subsets with at least 3 pairs.
So we start making the subsets with 8 pairs.
●With 8 ordered pairs…. we have just 1 subset , which is {(1,a),(1,b),(2,a,),(2,b),(3,a),(3,b),(4,a),(4,b)} (as such , every set is its own subset)
●similarly with 7 ordered pairs : the number of ways of choosing 7 pairs from the set of 8 pairs will be= 8C7 = 8!/(1! * 7!) = 8 subsets
● Now similarly with 6 pairs : we get 8C6
= 8!/(2! * 6!) = 28 subsets
● With 5 pairs : we get 8C5 = 8!/(3!*5!) = 56 subsets
● With 4 pairs : we get 8C4 = 8!/(4!*4!) = 70 subsets
●Now with 3 pairs: we get 8C3 = 8!/(5!*3!) = 56 subsets
Now, by adding all the above subsets , we get..
1 + 8 + 28 + 56 + 70 + 56 = 219 subsets . . . . Ans
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