If a committee of 3 people is to be selected from among 6 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?  Choose the most appropriate option:  A) 20  B) 40  C) 50  D) 80  E) 160
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.   12*10*8 / 6= 
<gadget id="calculator">12*10*8 / 6</gadget>
<output>160</output>
160. Answer 160.
<result>E</result>
If a committee of 3 people is to be selected from among 6 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?