There are 10 children in a company's day-care center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible? Choices.
A) 100  B) 90  C) 50  D) 45  E) 25
1. first child out of 10 - 10 possibilities   2. second child out of 9 - 9 possibilities. the total number of pairs = 9*10= 
<gadget id="calculator">9*10</gadget>
<output>90</output>
90   3. exclude xy yx cases:  90/2= 
<gadget id="calculator"> 90/2</gadget>
<output>45</output>
45   Second way   N=10C2=10!/(8!2!)=10*9/2= 
<gadget id="calculator">10*9/2</gadget>
<output>45</output>
45   ANS:45
<result>D</result>
There are 10 children in a company's day-care center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?