Jane and Thomas are among the 8 people from which a committee of 3 people is to be selected. How many different possible committees of 3 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?  Options: A) 24 B) 28 C) 32 D) 36 E) 40
The total number of ways to choose 3 people from 8 is 8C3 = 56.   The number of committees without Jane or Thomas is 6C3 = 20.   There are  56-20= 
<gadget id="calculator"> 56-20</gadget>
<output>36</output>
36 possible committees which include Jane and/or Thomas.   The answer is 36.
<result>D</result>
Jane and Thomas are among the 8 people from which a committee of 3 people is to be selected. How many different possible committees of 3 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?