first we can select a set for P and then for Q
6 cases will occur:
case 1: P is NULL set then Q can have 2^5 options.ways=2^5
case 2: p contains set with 1 element i.e. 5c1 and q will contain 2^4(we will remove the element given to p and count the no. of subsets formed with remaining 4 elements).ways=5*2^4
case 3: ways=5c2*2^3
case 5: ways=5c4*2^1
case 6: ways=5c5*2^0(complete set and null set).
total ways=sum all the six cases.
plz correct if wrong.
we can select atleast 1 from five different fruits
no. of ways=5c1+5c2+5c3+5c4+5c5=2^5-1
or we can say every fruit has 2 choices i.e 2^5 .We remove the choice when none is selected so total ways=2^5-1