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? Choose the correct choice from the following answers.
A) 100
B) 90
C) 50
D) 45 different pairs
E) 25
1. first child out of 0 - 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 different pairs
<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?