If a committee of 3 people is to be selected from among 9 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 correct choice.
A) 420 B) 540 C) 550 D) 672 E) 720
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.   18*16*14 / 6= 
<gadget id="calculator">18*16*14 / 6</gadget>
<output>672</output>
672. Answer 672.
<result>D</result>
If a committee of 3 people is to be selected from among 9 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?