Question 923981
I'll do the first one to get you started.


# 1


10 transistors
3 defective
7 nondefective



Number of ways to get exactly 2 defective


(3 C 2)*(7 C 1) = (3)*(7) = <font size=5 color="red">21</font>


There are <font size=5 color="red">21</font> ways to pick exactly 2 defective transistors (out of 3 selected transistors).


Side note: I'm using the <a href="http://www.mathwords.com/c/combination_formula.htm">combination formula</a> *[Tex \LARGE _{n}C_{r} = \frac{n!}{r!(n-r)!}]. Many calculators such as the TI84 have this formula built in (let me know if you need steps on how to use it).


Another side note: 3 C 2 represents picking exactly 2 bad transistors out of 3. 7 C 1 represents picking exactly 1 good transistor out of 7.