We are given that $n=11$ and $m=48$.
We want to form a committee consisting of teachers and students.
Let $k$ be the number of teachers in the committee and $l$ be the number of students in the committee.
The number of ways to choose $k$ teachers from $n=11$ teachers is given by the binomial coefficient:
$$ \binom{n}{k} = \binom{11}{k} = \frac{11!}{k!(11-k)!} $$
The number of ways to choose $l$ students from $m=48$ students is given by the binomial coefficient:
$$ \binom{m}{l} = \binom{48}{l} = \frac{48!}{l!(48-l)!} $$
The number of ways to form a committee with $k$ teachers and $l$ students is given by the product:
$$ \binom{11}{k} \binom{48}{l} $$
Since we can choose any number of teachers from 0 to 11 and any number of students from 0 to 48, we need to sum over all possible values of $k$ and $l$:
$$ \sum_{k=0}^{11} \sum_{l=0}^{48} \binom{11}{k} \binom{48}{l} $$
We can rewrite this sum as:
$$ \sum_{k=0}^{11} \binom{11}{k} \sum_{l=0}^{48} \binom{48}{l} $$
By the binomial theorem, we know that:
$$ \sum_{k=0}^{11} \binom{11}{k} = 2^{11} $$
$$ \sum_{l=0}^{48} \binom{48}{l} = 2^{48} $$
Therefore, the total number of different committees that can be formed is:
$$ 2^{11} \times 2^{48} = 2^{11+48} = 2^{59} $$
Thus, there are $2^{59}$ different committees that can be formed from 11 teachers and 48 students.
Final Answer: The final answer is $\boxed{2^{59}}$