Question

1. a variety pack of candy consists of nine bars each with a different flavor. if three bars of candy are chosen at random: how many different selections are possible? 80 a 87 b 84 c 36 d

Explanation:

Step1: Identify combination formula

Since order of selection does not matter, use the combination formula:
$$C(n,k)=\frac{n!}{k!(n-k)!}$$
where $n=9$ (total candy bars), $k=3$ (bars chosen)

Step2: Substitute values into formula

$$C(9,3)=\frac{9!}{3!(9-3)!}=\frac{9!}{3! \times 6!}$$

Step3: Simplify the factorial expression

$$\frac{9 \times 8 \times 7 \times 6!}{3 \times 2 \times 1 \times 6!}=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}$$

Step4: Calculate the final value

$$\frac{504}{6}=84$$

Answer:

C. 84