Question
Three out of seven students in the cafeteria line are chosen to answer survey questions.
How many different combinations of three students are possible?
7 C 3 = StartFraction 7 factorial Over (7 minus 3) factorial 3 factorial
7
35
70
210
How many different combinations of three students are possible?
7 C 3 = StartFraction 7 factorial Over (7 minus 3) factorial 3 factorial
7
35
70
210
Answers
GPT-4o mini
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:
1. Calculate \( 7! \):
\[
7! = 7 \times 6 \times 5 \times 4!
\]
So the \( 4! \) cancels out:
\[
\binom{7}{3} = \frac{7 \times 6 \times 5}{3!}
\]
2. Calculate \( 3! \):
\[
3! = 3 \times 2 \times 1 = 6
\]
3. Now substitute \( 3! \) back into the combination formula:
\[
\binom{7}{3} = \frac{7 \times 6 \times 5}{6}
\]
4. Now simplify:
\[
= 7 \times 6 \times 5 / 6 = 7 \times 5 = 35
\]
Thus, the number of different combinations of three students from seven is:
\[
\boxed{35}
\]
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
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:
1. Calculate \( 7! \):
\[
7! = 7 \times 6 \times 5 \times 4!
\]
So the \( 4! \) cancels out:
\[
\binom{7}{3} = \frac{7 \times 6 \times 5}{3!}
\]
2. Calculate \( 3! \):
\[
3! = 3 \times 2 \times 1 = 6
\]
3. Now substitute \( 3! \) back into the combination formula:
\[
\binom{7}{3} = \frac{7 \times 6 \times 5}{6}
\]
4. Now simplify:
\[
= 7 \times 6 \times 5 / 6 = 7 \times 5 = 35
\]
Thus, the number of different combinations of three students from seven is:
\[
\boxed{35}
\]