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}=\frac{7!}{(7 - 3)!3!})

Explanation:

Step1: Simplify the factorial expressions

$$\begin{align*} _7C_3&=\frac{7!}{(7 - 3)!3!}\\ &=\frac{7!}{4!3!}\\ &=\frac{7\times6\times5\times4!}{4!\times3\times2\times1} \end{align*}$$

Step2: Cancel out common terms and calculate

Cancel out the \(4!\) terms:

$$\begin{align*} \frac{7\times6\times5}{3\times2\times1}&=\frac{7\times6\times5}{6}\\ &=7\times5\\ & = 35 \end{align*}$$

Answer:

35