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Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is (1) 220 (2) 219 (3) 211 (4) 256

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is (1) 220 (2) 219 (3) 211 (4) 256

Grade:upto college level

2 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
Hello student,
As AXB has 8 elements
So number os subsets of AXB having 3 or more elements
=288C08C18C2= 256 − 1 − 8 − 28 = 219.
Thanks and Regards
Shaik Aasif
askIITians faculty
ankit singh
askIITians Faculty 614 Points
one year ago

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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