Permutations problem

[Darya]

We have to arrange 8 students in a row. One particular student "A" can only stand in the first three places and 2 other students, let's call them sisters, can stand in any order one after another, the rest of the students can stand anywhere anyhow.
So the number of places the "A" student can be in is 3, then we can arrange the sisters in 6*2 ways, others in 5! ways. So it's 4320 but the answer isn't correct. What am I doing wrong?

 

[Romsek]

There are 3 choices for A.

If A=1 then there are 6 slots for the 2 sisters
If A=2,3 there are 5 slots for the 2 sisters.

There are 2 ways to arrange the sisters in a slot

The remaining 5 students can be arranged in 5! ways.

Combining all this

[MATH]N= (2\cdot 5 + 6)\cdot 2! \cdot 5! = 3840[/MATH]

 

[Darya]

There are 3 choices for A.

If A=1 then there are 6 slots for the 2 sisters
If A=2,3 there are 5 slots for the 2 sisters.

There are 2 ways to arrange the sisters in a slot

The remaining 5 students can be arranged in 5! ways.

Combining all this

[MATH]N= (2\cdot 5 + 6)\cdot 2! \cdot 5! = 3840[/MATH]

Thank you so much!!!!! It's all clear