If 12 balls are thrown at random into 20 boxes, what is the probability that no box will receive more than one ball?
Please give only a hint, and not the full solution.
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 .
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 .
So,
I don't think that the answer is wrong. I think the model is wrong.
Why not assume that the boxes are identical and the balls are different?
OR, Why not assume that the boxes are identical and the balls are identical?
These are legitimate questions asked in counting theory.
How does one model these sorts of questions.