\( 11001 _ { 2 } - 110 _ { 2 } \)

To solve the binary subtraction \( 11001_2 - 110_2 \), the easiest method is to align the numbers and subtract bit by bit, borrowing when necessary. Here’s how it looks in binary: ``` 11001 - 110 -------- 10011 ``` Start from the rightmost bit: 1 - 0 = 1, next is 0 - 1 which requires borrowing (turning the next 0 into a 2). After borrowing, you get 10 - 1 = 1. The next bits give you 0 - 1 (again borrowing) which leaves you with 0 from the borrowed 1, resulting in 0. Lastly, you subtract the last two bits resulting in 1. Thus the result is \( 10011_2 \). Now, let's convert \( 10011_2 \) to decimal (base 10) to check our work: Starting from the right, you have \( 1 \times 2^0 + 1 \times 2^1 + 0 \times 2^2 + 0 \times 2^3 + 1 \times 2^4 \), which equals \( 1 + 2 + 0 + 0 + 16 = 19 \). The final answer is \( 10011_2 \) or \( 19 \) in decimal.