Question

an electronics store receives a shipment of 20 graphing calculators, 5 that are defective. four of the calculators are selected to be sent to a local high school.
(a) how many selections can be made?
(b) how many of these selections will contain no defective calculators?
(a) \\(\square\\) selections can be made.
(b) \\(\square\\) selections will contain no defective calculators.

Explanation:

Step1: Calculate total selections

Use combination formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, $n=20$, $k=4$.
$\binom{20}{4} = \frac{20×19×18×17}{4×3×2×1} = 4845$

Step2: Calculate non-defective selections

Non-defective calculators: $20-5=15$. Choose 4: $\binom{15}{4}$.
$\binom{15}{4} = \frac{15×14×13×12}{4×3×2×1} = 1365$

Answer:

(A) 4845
(B) 1365