Originally Posted by
alexmahone My solution:
No. of favourable arrangements:
The 1st ball can be put into any of the 20 boxes, the 2nd ball can can be put into any of the other 19 boxes and so on. So, the number of ways to put 12 balls into 20 boxes so that no box receives more than one ball is $\displaystyle 20*19*\cdots*9=\frac{12!}{8!}$.
Total no. of arrangements:
The 1st ball can be put into any of the 20 boxes, the 2nd ball can be put into any of the 20 boxes and so on. So, the total number of ways to put 12 balls into 20 boxes is $\displaystyle 20^{12}$.
So, $\displaystyle P=\frac{12!}{8!20^{12}}$