Question

Let A and B be two set containing 2 element and 4 element respectively .the number of subsets of A×B having 3 or more elements is

Answer 1

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

Answer 2

Answer:

219 is your answer.

Step-by-step explanation:

Given Problem:

Let A and B be two set containing 2 element and 4 element respectively .the number of subsets of A×B having 3 or more elements is.

Solution:

A = {x,y}

B = {a,b,c,d}

A × B having 2 × 4 = 8 elements.

∴ Total substance of A × B is 2^8 = 256

$\tt =256-\bigg(1\;null\;set+8\;single\;tonset+\;_8C_2\;having\;2\;elements\bigg)$

$\sf =256-1-8-28=219$