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?
$_7c_3 = \frac{7!}{(7 - 3)!3!}$
\bigcirc 7
\bigcirc 35
\bigcirc 70
\bigcirc 210

Explanation:

Step1: Recall combination formula

The number of ways to choose \( k \) items from \( n \) items is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). Here, \( n=7 \), \( k=3 \).

Step2: Calculate factorials

\( 7! = 7×6×5×4! \), \( 3! = 6 \), \( (7-3)! = 4! \).

Step3: Simplify the formula

\( \binom{7}{3} = \frac{7×6×5×4!}{3!×4!} = \frac{7×6×5}{3×2×1} = 35 \).

Answer:

35