Let's first visualize the set A (set of distinct integers).
n -> number of elements (integers) in the set A, overall
k -> number of odd integers in the set A
n - k -> number of even integers in the set A
A -> { (.....k odd integers.....) (.....n-k even integers.....) }Number of subsets of the set = \(2^n\) (this includes the null set i.e., nC0).
Number of subsets of A that contain no even integers = \(2^k\) (we have to ignore the even numbers - from the odd numbers, how many sets can be created is what will yield this).
Hence,
The number of subsets of A that contain at least one even integer = \(2^n\) - \(2^k\)
Given:300 < \(2^n\) - \(2^k\) < 400
Analyzing this further to see what values n and k can take:\(2^n\) > 300 + \(2^k\)
=> n >=9
Can n be 10?\(2^n\) < 400 + \(2^k\)
1024 < 400 + \(2^k\)
\(2^k\) > 624
which means k>=10.
But this is a problem. If n = 10, k has to be lesser than 10. k>=10 is impossible.
In other words, we can see that n cannot be 10.
We can even check whether n = 11 works. Then again, we observe that k has to be >=11. Which is impossible.
Key Insight: We already established that n>=9. But we can also see that n cannot be >9. Hence, n = 9.
So, given that n = 9,
512 > 300 + \(2^k\)
=> k <=7
512 < 400 + 2^k
=> k >=7
=> k = 7
Thus, in set A, we have 9 integers in total, out of which we have 7 odd numbers and 2 even numbers.
How many 3-element subsets of A are such that the sum of the 3 integers in the subset is even?
The only way to get an even sum is if 2 odd numbers and 1 even number are picked.
How many such subsets can be created?
7C2 x 2C1 = 42. Choice D.
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Harsha