<PRE><B>Question 1120042</B>
&lt;br&gt;
The answer by the other tutor is obtained by choosing the three members of one team, then the three number of the second team, and then the three members of the third team, leaving the last person to be the field judge.&lt;br&gt;
It would seem the answer would be different if you chose the three teams and the field judge in a different order.  But it turns out that is not the case.&lt;br&gt;
(1) Choose the field judge last:
C(10,3)*C(7,3)*C(4,3)*C(1,1) = 120*35*4*1 = 16800&lt;br&gt;
(2) Choose the field judge after two teams:
C(10,3)*C(7,3)*C(4,1)*C(3,3) = 120*35*4*1  = 16800&lt;br&gt;
(3) Choose the field judge after just one team:
C(10,3)*C(7,1)*C(6,3)*C(3,3) = 120*7*20*1 = 16800&lt;br&gt;
(4) Choose the field judge first:
C(10,1)*C(9,3)*C(6,3)*C(3,3) = 10*84*20*1 = 16800&lt;br&gt;
Here is another way to get the answer without having to worry about the order in which the teams are formed.&lt;br&gt;
Imagine each person getting assigned to either team A, team B, team C, or the field judge F.  Then&lt;br&gt;
AAABBBCCCF  would mean team A consists of players 1, 2, and 3; team B of players 4, 5, and 6; team C of players 7, 8,and 9; with player 10 the field judge.&lt;br&gt;
AABCCBAFCB  would mean team A consists of players 1, 2, and 7; team B of players 3, 6, and 10, team C of players 4, 5, and 9, with player 8 the field judge.&lt;br&gt;
Each distinct arrangements of the letters AAABBBCCCF represents one of the possible ways the players can be divided into three teams of 3 and a field judge.  By a well known counting principle, the number of different ways to do that is&lt;br&gt;
{{{(10!)/((3!)(3!)(3!)(1!)) = 3628800/216 = 16800}}}</PRE>