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Simple probability question

Alexmahone

Active member
Jan 26, 2012
268
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

Well-known member
Feb 13, 2012
1,704

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,682
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

Active member
Jan 26, 2012
268
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 [TEX]20*19*\cdots*9=\frac{20!}{8!}[/TEX].

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 [TEX]20^{12}[/TEX].

So, [TEX]P=\frac{20!}{8!20^{12}}[/TEX]