Problem: Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$

Solution: We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
$\textbf{Case 1: (1,2)}$
If we have (1,2) as our pair, we are left with the numbers from 3-10 as elements that can be added to our subset. So, we must compute how many ways we can pick these numbers so that the set has no consecutive numbers other than (1,2). Our first option is to pick no more numbers, giving us $8 \choose {0}$ . We can also pick one number, giving us $7 \choose {1}$ because 3 cannot be picked. Another choice is to pick two numbers and in order to make sure they are not consecutive we must fix one number in between them, giving us $6 \choose {2}$ . This pattern continues for each amount of numbers, yielding $5 \choose {3}$ for 3 numbers and $4 \choose {4}$ for four numbers. Adding these up, we have $8 \choose {0}$ $7 \choose {1}$ $6 \choose {2}$ $5 \choose {3}$ $4 \choose {4}$ $\textbf{34}$
$\textbf{Case 2: (2,3)}$
If we have (2,3) as our pair, everything works the same as with (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we have $7 \choose {0}$ $6 \choose {1}$ $5 \choose {2}$ $4 \choose {3}$ $\textbf{21}$
$\textbf{Case 3: (3,4)}$
This case remains pretty much the same except we now have an option of whether or not to include 1. If we want to represent this like we have with our other choices, we would say $2 \choose {0}$ for choosing no numbers and $1 \choose {1}$ for choosing 1, leaving us with $2 \choose {0}$ $1 \choose {1}$ = 2 choices (either including the number 1 in our subset or not including it). As far as the numbers from 5-10, our pattern from previous cases still holds. We have $6 \choose {0}$ $5 \choose {1}$ $4 \choose {2}$ $3 \choose {3}$ = 13. With 2 choices on one side and 13 choices on the other side, we have $2\cdot13$ $\textbf{26}$ combinations in all.
$\textbf{Case 4: (4,5)}$
Following the patterns we have already created in our previous cases, for the numbers 1-3 we have $3 \choose {0}$ $2 \choose {1}$ = 3 choices (1, 2, or neither) and for the numbers 6-10 we have $5 \choose {0}$ $4 \choose {1}$ $3 \choose {2}$ = 8 choices. With 3 choices on one side and 8 choices on the other side, we have $3\cdot8$ $\textbf{24}$ combinations in all.
$\textbf{Case 5: (5,6)}$
Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have $4 \choose {0}$ $3 \choose {1}$ $2 \choose {2}$ = 5 choices and for the numbers 5-10 we have the same $4 \choose {0}$ $3 \choose {1}$ $2 \choose {2}$ = 5 choices. $5\cdot5$ $\textbf{25}$ combinations in all.
$\textbf{Rest of the cases}$
By symmetry, the case with (6,7) will act the same as case 4 with (4,5). This goes the same for (7,8) and case 3, (8.9) and case 2, and (9,10) and case 1.
Now, we simply add up all of the possibilities for each case to get our final answer. 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 = $\boxed{235}$

Answer:  235