Simple probability question

[alexmahone]

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.

 

[chisigma]

The correct approach to the problem requires [probably...] the multinomial distribution...

http://mathworld.wolfram.com/MultinomialDistribution.html

Kind regards

$\chi$ $\sigma$

 

[anemone]

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.

 

[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 $$20*19*\cdots*9=\frac{20!}{8!}$$.

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, $$P=\frac{20!}{8!20^{12}}$$

 

[CellCoree]

The key to solving this problem is to think about the total number of possible outcomes and the number of favorable outcomes. Consider the fact that each ball can only go into one box, and that the order in which the balls are thrown does not matter. How many different ways can 12 balls be distributed into 20 boxes without any box receiving more than one ball?