Problem: Let $B$ be the set of all binary integers that can be written using exactly 5 zeros and 8 ones where leading zeros are allowed. If all possible subtractions are performed in which one element of $B$ is subtracted from another, find the number of times the answer 1 is obtained. Solution: It suffices to count the $x \in B$ such that $x+1 \in B$. If $x \in B$ is even, then it ends with a $0$ and $x+1$ is odd and ends with a $1$. But then $x+1$ would have $9$ ones and $4$ zeros meaning $x+1$ is not in the set to begin with. Thus, $x$ ends with a $1$. Notice that if $x$ ends with $11$, then $x+1$ will have strictly fewer ones than the requirement. We need to make sure that adding $1$ to $x$ preserves the number of $1s and $0s so $x$ ends with $01$. That is $x \in B$ and $x+1 \in B$ if and only if $x$ ends with $01$. So, if we count all binary words with $4$ zeros and $7$ ones, we are done, as we just append $01$ to the end of it. The number of such binary strings is $\binom{11}{4}$.