QJEEM19Ma68e2: JEE Main 2019 Mathematics - Class 12 JEEM19Ma68e2 Solution: Let  S = {1, 2, . . . . . ., 20}. A subset B of S is…

We can solve this problem by counting the number of "nice" subsets in the set S = {1, 2, $\ldots$ , 20 }, and then dividing that number by the total number of possible subsets of S.

The sum of all elements in S is :

1 + 2 + $\ldots$ + 20 = $\frac{{20 \times 21}}{2}$ = 210

Since a "nice" subset must sum to 203, the elements not in the subset must sum to 210 - 203 = 7.

Now we need to find the ways to make the sum of 7 using the elements of S. The combinations are :

1. 1. 7
2. 2. 1 + 6
3. 3. 2 + 5
4. 4. 3 + 4
5. 5. 1 + 2 + 4
6. 6. 1 + 3 + 3(This doesn't work since 3 is repeated)
7. 7. 2 + 2 + 3(This doesn't work since 2 is repeated)

So, there are 5 "nice" subsets.

Since the set S has 20 elements, there are $2^{20}$ possible subsets (including the empty set and the set itself). The probability of randomly choosing a "nice" subset is therefore :