If $$2 \leq |x - 1| \times  |y + 3| \leq 5$$ and both $$x$$ and $$y$$ are negative integers, find the number of possible combinations of $$x$$ and $$y$$.

2 ≤ |x - 1| × |y + 3| ≤ 5
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x - 1| will be 2 and the maximum value of |x - 1| will be 5 as per the question.

When, |x - 1| = 2, |y + 3| can be either 1 or 2
So, for x =  -1, y can be - 4 or - 2 or - 5 or -1.
Thus, we get 4 pairs of (x, y) 

When |x - 1| = 3, |y + 3| can be 1 only
So, for x = - 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x - 1| = 4, |y + 3| can be 1 only
So, for x = - 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x - 1| = 5, |y + 3| can be 1 only
So, for x = - 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

Therefore, we get a total of 10 pairs of the values of (x, y)
Hence, option E is the correct answer.