If a committee of 3 people is to be selected from among 8 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?	Choices:
A) 120	B) 240	C) 250	D) 380	E) 448
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.   16*14*12 / 6= 
<gadget id="calculator">16*14*12 / 6</gadget>
<output>448</output>
448. Answer 448.
<result>E</result>
If a committee of 3 people is to be selected from among 8 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?