To determine how many different combinations of three students can be chosen from a group of seven students, we can use the combination formula [tex]\( \binom{n}{r} \)[/tex], which is defined as:
[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]
In this problem, we need to find the number of combinations for choosing 3 students out of 7. This means [tex]\( n = 7 \)[/tex] and [tex]\( r = 3 \)[/tex].
Using the combination formula:
[tex]\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} \][/tex]
Simplifying the factorial expressions:
- [tex]\( 7! \)[/tex] represents the factorial of 7, which is [tex]\( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)[/tex]
- [tex]\( 3! \)[/tex] represents the factorial of 3, which is [tex]\( 3 \times 2 \times 1 \)[/tex]
- [tex]\( (7-3)! \)[/tex] represents the factorial of 4, which is [tex]\( 4 \times 3 \times 2 \times 1 \)[/tex]
Plugging these into the formula:
[tex]\[ \binom{7}{3} = \frac{7!}{3! \cdot 4!} \][/tex]
Given our known correct result from performing the calculation, we know:
[tex]\[ \binom{7}{3} = 35 \][/tex]
Thus, the number of different combinations of three students that are possible is [tex]\( \boxed{35} \)[/tex].
