Simple probability and statistics problem If there are 12 football teams in a league, how many different bets can you make if you bet on 3 first teams and 2 that will get kicked out of the league?
The solutions says its (12*11*10*9*8)/2. 
Why do you divide by two? Is it because the twos that get kicked out, their order doesnt matter so there are half of their variations? What if the bet was on 3 first and 3 that get kicked out?
 A: The question is not very clear. Does the bet look like "I bet the the first command will be $A$, the second command will be $B$, the third command will be $C$, and kicked off commands will be $D$ and $E$"? Or "I bet the the first three commands will be $A$, $B$ and $C$, the kicked off commands will be $E$ and $D$"? These are different kinds of bets. F.e. if the actual final chart is ($B$, $C$, $A$, ... $E$, $D$) the first bet looses, but the second bet wins. And if the final chart is ($A$, $B$, $C$, ... $E$, $D$) both bets win.
Based on the answer provided the question was: how many different bets like "I bet the the first command will be $A$, the second command will be $B$, the third command will be $C$, and kicked off commands will be $D$ and $E$" are possible?
To answer this question let's write down on a separate sheets all the possible final charts with filled first three and last two commands. Each chart will look like ($A$, $B$, $C$, ... $E$, $D$). There will be $12*11*10*9*8$ sheets of paper. Based on your comment I guess this is clear.
But charts ($A$, $B$, $C$, ... $E$, $D$) and ($A$, $B$, $C$, ... $D$, $E$) correspond to the same bet! And only two sheets correspond to this bet. So we group all our sheets in pairs and get the the number of possible bets is $12*11*10*9*8 / 2$. That's where the "divide by 2" comes from.
Now if the question was about 3 commands to be kicked off, there will be different sheets of paper with charts like ($A$, $B$, $C$, ... $E$, $D$, $F$). And there will be more sheets corresponding to the same bet. Actually you can permutate the last three commands in every possible way, and the chart will still correspond to the same bet. Number of such permutations is $3!$, so you will have to divide the number of sheets by 6.
