To determine how many different combinations of three students can be chosen from seven, you use the combination formula, which is given by:
[tex]\[
{ }_n C_k=\frac{n!}{(n - k)! \cdot k!}
\][/tex]
For our specific problem, [tex]\( n = 7 \)[/tex] and [tex]\( k = 3 \)[/tex]. Plugging these values into the combination formula, we get:
[tex]\[
{ }_7 C_3=\frac{7!}{(7-3)! \cdot 3!}
\][/tex]
First, let's simplify the factorials in the denominator:
[tex]\[
(7-3)! = 4!
\][/tex]
[tex]\[
3! = 3 \times 2 \times 1 = 6
\][/tex]
Next, compute [tex]\( 7! \)[/tex]:
[tex]\[
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
\][/tex]
Now, substitute these back into the formula:
[tex]\[
{ }_7 C_3=\frac{7!}{4! \cdot 3!} = \frac{5040}{4! \cdot 6}
\][/tex]
Compute [tex]\( 4! \)[/tex]:
[tex]\[
4! = 4 \times 3 \times 2 \times 1 = 24
\][/tex]
Then, we have:
[tex]\[
{ }_7 C_3=\frac{5040}{24 \cdot 6} = \frac{5040}{144}
\][/tex]
Finally, performing the division:
[tex]\[
\frac{5040}{144} = 35
\][/tex]
Therefore, the number of different combinations of three students from seven is:
[tex]\[
\boxed{35}
\][/tex]
