the number of ways 7children can be arranged in a line, if three of them must be separated, is: ( )
+129852 I got the same answer as Melody, but with a slightly different approach.....
Here are all the "sets" ot the positions that the "separated" children can occupy :
1 3 5 2 4 6
1 3 6 2 4 7
1 3 7 2 5 7
1 4 6 3 5 7
1 4 7
1 5 7
And for each set, the "separarted" children can be arranged in 3! ways
And for each of these arrangements, the other four children can be arranged in 4! ways in the other 4 positions.........so
10 sets * 3! arrangements of the separated set * 4! arrangements of the other children = 1440 total arrangements.......
+1316 There are 7! ways to arrange all the children.
There are 3! way to arrange 3 children.
There can be 3 pairs of children from the 3 children.
There are 2! +2! +2! ways of arranging the pairs of children.
So the answer is 7! - 3! - 2! - 2! - 2! ways to arrange the children with none of the 3 next to each other.
+118677 Good try Dragonlance,
Put the four non-problem children in a row first. There are 4! ways to do this. 4!=24
Now put the other childen in the row. One child can go between each of the original 4 and/or one child can go at the beginning and/or the end of the line.
So there are 5 positions that the 3 can occupy. There are 5P3=60 ways that this can be done.
So altogether there are 24*60 = 1440 waysthe children can be put in the line. 
+129852 I got the same answer as Melody, but with a slightly different approach.....
Here are all the "sets" ot the positions that the "separated" children can occupy :
1 3 5 2 4 6
1 3 6 2 4 7
1 3 7 2 5 7
1 4 6 3 5 7
1 4 7
1 5 7
And for each set, the "separarted" children can be arranged in 3! ways
And for each of these arrangements, the other four children can be arranged in 4! ways in the other 4 positions.........so
10 sets * 3! arrangements of the separated set * 4! arrangements of the other children = 1440 total arrangements.......
+118677 Hi Chris,
That is an interesting alternate solution.
I wonder how easy it would be for either of us if the question were harder.
Say 3 had to be seperated and there were 25 others to go into the line as well.
How hard is this one?
It is too late for me to think about it now. Maybe you might like to consider it though :)
+118677 Hey Chris, do I have to answer my own question :/ LOL
I wonder how easy it would be for either of us if the question were harder.
Say 3 had to be seperated and there were 25 others to go into the line as well.
How hard is this one?
It is too late for me to think about it now. Maybe you might like to consider it though :)
This is what I reckon.
There are 25! ways the non trouble makers can be put in a line.
Then there are 26 spots where the other 3 can be slipped in so that is 26P3 = 26*25*24 = 15600
So there would be 15600*25! ways.
And this is about 2.4*10^(29) and that is a HUGE number of ways :))
