Answer - Option (B)
Dividing the set into two sets for convenience without loss of generality.
$S_1 = \{ 10, 11, 12, 13, ..., 49 \}$ and $S_2 = \{50,51,52,53, ..., 99 \}$
Here, No. of elements in $S_1 = 40$ and No. of elements in $S_2 = 50$
Now, to make $x + y \geq 100$
we can choose either any 2 numbers from $S_2$ [Part A] or one number from $S_1$ and one number from $S_2$ [Part B].
Since ordered pairs are not asked, pairs $(a,b)$ and $(b,a)$ will be same, hence
[Part A]
Total no. of choices = ways of selecting $2$ distinct numbers from set $S_2$= $^{50} C_2 = \frac{50 \times 49}{2}$
but the choices like $(50,50)$, $51,51$ are missing above. So, no. of choices such that $x$ and $y$ are equal are $= 50$
[Part B]
Now this part is tricky. We have to choose $1$ number from $S_1$ and $1$ number from $S_2$.
For numbers $\{90, 91, 92, ..., 99 \}$ we can choose any number from $S_1$ and have $x + y \geq 100$
So, for these $10$ numbers, we have 40 choices.
So, total no. of choices $ = 10 \times 40 = 400$
Now for $89$, we can choose everything from $S_1$ except number $10$ since that would fail the requirement of $x+y \geq 100$.
So,
for $89$ we have $40-1 = 39$ choices
Similarly, for $88$ we have $40 -2 = 38$ choices since $\{10, 11 \}$ from $S_1$ will fail our criteria of sum greater than 100.
Similarly, for $87$, we have $40-3 = 37$ choices
...
Similarly, for $51$, we have $1$ choice (which is $49$)
Similarly, for $50$, we have $0$ choices.
Hence total no. of choices $ = 0 + 1 + 2+ 3+ 4+ ... + 39 = \frac{(39)(40)}{2} = 780$
Hence, Answer $= 1225 +50+780 + 400 = 2455$