How many ways are there to partition $7$ students into the groups of $2$ or $3$?

$ 
\textbf{(A)}\ 70
\qquad\textbf{(B)}\ 105
\qquad\textbf{(C)}\ 210
\qquad\textbf{(D)}\ 280
\qquad\textbf{(E)}\ 630
$
To solve the problem of partitioning 7 students into groups of 2 or 3, we need to consider the possible groupings and count the number of ways to form these groups.

1. **Identify the possible groupings:**
   - One possible way to partition 7 students is to have one group of 3 students and two groups of 2 students.

2. **Calculate the number of ways to choose the group of 3 students:**
   - The number of ways to choose 3 students out of 7 is given by the binomial coefficient:
     \[
     \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
     \]

3. **Calculate the number of ways to partition the remaining 4 students into two groups of 2:**
   - After choosing the group of 3 students, we have 4 students left. The number of ways to choose 2 students out of these 4 is given by:
     \[
     \binom{4}{2} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6
     \]
   - However, this counts each pair twice (e.g., choosing {a, b} and {c, d} is the same as choosing {c, d} and {a, b}). Therefore, we need to divide by 2 to account for this overcounting:
     \[
     \frac{\binom{4}{2}}{2} = \frac{6}{2} = 3
     \]

4. **Combine the results:**
   - The total number of ways to partition the 7 students is the product of the number of ways to choose the group of 3 students and the number of ways to partition the remaining 4 students into two groups of 2:
     \[
     35 \times 3 = 105
     \]

The final answer is \(\boxed{105}\).