If a committee of 3 people is to be selected from among 4 married couples so that the committee does not include two people who are married to each other, how many such committees are possible? Pick:
A) 20	B) 32	C) 54	D) 80	E) 120
You now have your Committee of 3 people. But you've selected them in an order, when order doesn't matter.   Selection #1 = Jack   Selection #2 = Susan   Selection #3 = Mary   Assume this is a good committee where Jack is not married to Susan or Mary and Susan and Mary are not married. The above committee will be the same as the following committee:   Selection #1 = Susan   Selection #2 = Mary   Selection #3 = Jack   To account for the fact that the order does not matter, we must divide out the number of ways a single committee can be the same members but in different order. 3 places = 3!, 3*2*1 = 6.   8*6*4 / 6= 
<gadget id="calculator">8*6*4 / 6</gadget>
<output>32</output>
32. Answer 32.
<result>B</result>
If a committee of 3 people is to be selected from among 4 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?