Carmen selects four different numbers from the set $\{1, 2, 3, 4, 5, 6, 7\}$ whose sum is 11. If $l$ is the largest of these four numbers, what is the value of $l$?
1. We need to find four different numbers from the set $\{1, 2, 3, 4, 5, 6, 7\}$ whose sum is 11. Let these numbers be $a, b, c,$ and $l$ where $a < b < c < l$.
2. The sum of these four numbers is given by:
   \[
   a + b + c + l = 11
   \]
3. To find the possible values of $l$, we start by considering the smallest possible values for $a, b,$ and $c$.
4. The smallest three numbers in the set are $1, 2,$ and $3$. Let's assume $a = 1$, $b = 2$, and $c = 3$. Then we have:
   \[
   1 + 2 + 3 + l = 11
   \]
   Simplifying this, we get:
   \[
   6 + l = 11 \implies l = 5
   \]
5. We need to verify that there are no other combinations of four different numbers that sum to 11. Let's consider other possible values for $a, b,$ and $c$:
   - If $a = 1$, $b = 2$, and $c = 4$, then:
     \[
     1 + 2 + 4 + l = 11 \implies 7 + l = 11 \implies l = 4
     \]
     But $l$ must be different from $a, b,$ and $c$, so this is not valid.
   - If $a = 1$, $b = 3$, and $c = 4$, then:
     \[
     1 + 3 + 4 + l = 11 \implies 8 + l = 11 \implies l = 3
     \]
     Again, $l$ must be different from $a, b,$ and $c$, so this is not valid.
   - If $a = 2$, $b = 3$, and $c = 4$, then:
     \[
     2 + 3 + 4 + l = 11 \implies 9 + l = 11 \implies l = 2
     \]
     Again, $l$ must be different from $a, b,$ and $c$, so this is not valid.
6. Therefore, the only valid combination is $a = 1$, $b = 2$, $c = 3$, and $l = 5$.

The final answer is $\boxed{5}$