Bali pairs conceptual doubt

#1

i am not understanding what mentor wants to convey .can u explain in detail

#2

@neha_153
There are some pairs given to you. Pairs may/may not contain same element.(Don’t get confused because of “shoes”)
You have to select one element from each pair.
Sum of all elements you pick should be odd.
For given test case,these are possible:
1+3 (even,invalid)
1+4 (odd,valid)
2+3 (odd,valid)
2+4 (even,invalid)
Hence answer is 2.
dont forget to hit like and mark resolved if cleared :smiley:

#3

i want to know the last case where (odd even pair)!=0 then whats the scenario

#4

@neha_153 starting from 7.45 third case is explained refer that.
in short we assume we have taken n-1 numbers in 2 to the power n-1 ways. so at the last odd even pair we can only select one acc to the prev n-1 sum thats why ans is 2 pow n-1