To find the number of different combinations of three students that can be chosen from a group of seven, we use the combination formula:
[tex]\[
{ }_n C_r = \frac{n!}{(n-r)! \cdot r!}
\][/tex]
For this particular problem, we have [tex]\( n = 7 \)[/tex] and [tex]\( r = 3 \)[/tex]. Plugging these values into the combination formula, we get:
[tex]\[
{ }_7 C_3 = \frac{7!}{(7-3)! \cdot 3!}
\][/tex]
First, calculate the factorials:
- [tex]\( 7! \)[/tex] (7 factorial) is the product of all positive integers up to 7:
[tex]\[
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
\][/tex]
- [tex]\( (7-3)! = 4! \)[/tex] (4 factorial) is the product of all positive integers up to 4:
[tex]\[
4! = 4 \times 3 \times 2 \times 1 = 24
\][/tex]
- [tex]\( 3! \)[/tex] (3 factorial) is the product of all positive integers up to 3:
[tex]\[
3! = 3 \times 2 \times 1 = 6
\][/tex]
Now, substitute these factorials back into the combination formula:
[tex]\[
{ }_7 C_3 = \frac{5040}{24 \cdot 6}
\][/tex]
Calculate the denominator:
[tex]\[
24 \cdot 6 = 144
\][/tex]
So, we have:
[tex]\[
{ }_7 C_3 = \frac{5040}{144}
\][/tex]
Finally, divide 5040 by 144:
[tex]\[
\frac{5040}{144} = 35
\][/tex]
Therefore, the number of different combinations of three students from a group of seven is:
[tex]\[
\boxed{35}
\][/tex]
