Question

An electronics store receives a shipment of 20 graphing calculators, including 3 that aredefective. Four of the calculators are selected to be sent to a local high school. Howmany of these selections will contain no defective calculators?

Answer 1

Given:

Total calculatore = 20

Defective =3

Total non defective calculatore is:

$\begin{gathered} =20-3 \\ =17 \end{gathered}$

For four calculatore:

$\begin{gathered} ^{17}C_4 \\ =\frac{17!}{4!(17-4)!} \\ =\frac{17\times16\times15\times14\times13!}{13!\times4\times3\times2\times1} \\ =2380 \end{gathered}$

So total 2380 selection will contain no defective calculators.