Here in this video it is saying that because we have $960$ distinct starting positions therefore it is known as chess $960$.
I tried to solve this problem by myself but I failed.
I have two rooks, two bishops, two knights, a queen and the king. Using this I have to fill all the eight squares on the first rank of the chess board. In addition there are two constraints that, (i) both the bishops must be placed on different colors and (ii) the king will sit between the rooks.
Starting with the queen, queen can take any of the eigth positions and both the knights can take any of the seven and six positions respectively. Now definitely one of the bishops have three possible positions to take and the other bishop has two possible positions to take. So so far we have $8\times 7\times 6\times3 \times 2=2016$ and this is more than $960$. Surely I'm making a mistake. As always combinatorics is a headache for me.
Please help me out.
What you have isn't quite right, for two reasons. First, the two knights are the same, so it doesn't matter whether you put the first knight on square A and the second on square B, or the first on B and the second on A. So there are only $8\times7\times6/2$ ways to place the queen and knights. Secondly, if you place the queen and knights on three squares of the same colour then there are only $4\times 1$ ways to place the bishops, not $3\times 2$ (one of them has to go on the fourth square of that colour).
It's easier to place the bishops first, in $4\times4$ ways. Now there are $6\times5\times4/2$ ways to place the queen and the knights. Once you've done this, there's only one way to place the king and rooks (the king in the middle of the three remaining squares). This gives $4\times4\times6\times5\times4/2=960$.