A stock of 12 DVD players contains two defective ones. In how many ways can one select 5 of these for inspection such that:

a. Only one of the defective DVD players is included.
b. Both defective DVD players are included.
c. none of the defective DVD players is included.

Question

A stock of 12 DVD players contains two defective ones. In how many ways can one select 5 of these for inspection such that:

a. Only one of the defective DVD players is included.
b. Both defective DVD players are included.
c. none of the defective DVD players is included.

sirm3d · Answer

a. choose one defective disc and one non-defective disc.
for a defective disc, it can be selected in $\displaystyle \binom{2}{1}$ ways.
for a non-defective disc, it can be done in $\displaystyle \binom{10}{4}$ ways.

anonymous · Answer

sirm3d is correct, but he might as well finish it off

anonymous · Answer

$$\frac{\binom{2}{1}\binom{10}{4}}{\binom{12}{5}}$$ finishes it
others are similar

anonymous · Answer

why divide by $$\left(\begin{matrix}12 \ 5\end{matrix}\right)$$

sirm3d · Answer

i prefer to see his work. seeing him solve the problem is better than showing him all the solution.

sirm3d · Answer

there is no need to divide. it is not a probability question, just a counting problem.

anonymous · Answer

letter b is (2C2)(10C3) right?

anonymous · Answer

and letter c is (2C0)(10C5) right?

sirm3d · Answer

right you are!

anonymous · Answer

thanks

sirm3d · Answer

yw.