To calculate the number of different combinations of three students from a group of seven, we use the combination formula, which is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
In this case, \( n = 7 \) and \( r = 3 \). Therefore, we need to calculate:
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!} \]
We can simplify this calculation as follows:
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Calculate \( 7! \): \[ 7! = 7 \times 6 \times 5 \times 4! \] So the \( 4! \) cancels out: \[ \binom{7}{3} = \frac{7 \times 6 \times 5}{3!} \]
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Calculate \( 3! \): \[ 3! = 3 \times 2 \times 1 = 6 \]
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Now substitute \( 3! \) back into the combination formula: \[ \binom{7}{3} = \frac{7 \times 6 \times 5}{6} \]
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Now simplify: \[ = 7 \times 6 \times 5 / 6 = 7 \times 5 = 35 \]
Thus, the number of different combinations of three students from seven is:
\[ \boxed{35} \]