Given $20$ balls labeled $1$ to $20$, and $10$ bins. How many permutations (order matters) are there if we put $2$ balls in each bins?
My attempt:
Take the first bin, there are ${^{20}\text{P}_2}$ ways of choosing $2$ balls to go into it. Then, there are ${^{18}\text{P}_2}$ ways of choosing. And so on.
So in total, there are $\Pi^{10}_{n=1} \,{^{n}\text{P}_2}$ ways.
Question:
Is this correct?