MHB Simple probability question

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The probability of throwing 12 balls into 20 boxes without any box receiving more than one ball can be approached using the multinomial distribution. The number of favorable arrangements is calculated by considering that the first ball can go into any of the 20 boxes, the second into any of the remaining 19, and so forth, resulting in 20*19*...*9, or 20!/(8!). The total arrangements allow each ball to go into any of the 20 boxes, leading to 20^12 possible outcomes. The final probability is expressed as P = (20!/(8!))/(20^12).
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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.
 
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Alexmahone said:
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.

I believe we share the same attitude: Seeing the full solution is like, killing our imagination. :p

Hint:
The question can be rewritten in another way so that it's very easy for us to apply the formula.

 
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 20*19*\cdots*9=\frac{20!}{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 20^{12}.

So, P=\frac{20!}{8!20^{12}}
 
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