Question

S consists of a set of numbers from -50 to 50 (zero inclusive), we have to pick a set K such that K consists of {A, B, C} such that the sum of numbers in set K is equal to zero. How many such sets are possible?

• A. 1250
• B. 650
• C. 1300
• D. 1220

Answer 1

The correct option is C

1300

We can have either 2 +ve numbers and 1 -ve number or 2 -ve numbers and 1 +ve number.
Let's consider the case when we have 1 -ve and 2 +ve numbers.
For the -ve number we can assume it to be -50, the +ve numbers have to be chosen in pairs such that 2 of them add upto 50. i.e. (0,50),(1,49).. We see that 25 such combinations are possible. (Note: Here only 25 will not have a pair)
Similarly for -49, 25 such combinations will be possible. (Note: Here -50 will be left out)
Based on same logic we can extrapolate: Hence the total number of combinations possible will be 2(25 + 24 + 23 +.... +1).
As we can also have 2 -ve numbers and 1 +ve number, the same number of combinations will again be possible.

Hence, the total number of solutions will be $2\times 2\times \frac{n(n+1)}{2}$; where n = 25. Thus, the total number of ways will be 1300.