My Math Forum Exercise with boxes and balls.

 October 20th, 2013, 10:47 AM #1 Member   Joined: Apr 2013 Posts: 52 Thanks: 0 Exercise with boxes and balls. We have n boxes and m numbered (distinct) balls.With how many ways can we put the balls in the boxes so that no box contains more than one ball, if (a) n> = m, and (b) n
October 20th, 2013, 05:49 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: Exercise with boxes and balls.

Hello, evinda!

Part (b) is silly, isn't it?
If there are fewer boxes than there are balls,
at least one box will have more than one ball.

Quote:
 $\text{W\!e have }n\text{ boxes and }m\text{ numbered (distinct) balls.} \text{In how many ways can we put the balls in the boxes} \text{so that no box contains more than one ball if:} (a)\;n\,\ge\,m$

$\text{Select }m\text{ of the }n\text{ boxes.}
\text{There are: }\,{n\choose m}\text{ ways.}$

$\text{Place a numbered ball into each of the }m\text{ boxes.}
\text{There are }\,m!\text{ possible placements.}$

$\text{Therefore, there are: }\:{n\choose m}\,\cdot\,m! \;=\;\frac{n!}{m!\,(n-m)!}\,\cdot\,m! \;=\;\frac{n!}{(n-m)!}$ $\text{ ways.}$

Quote:
 Answer the same questions when the balls are identical (indistinguishable).

$\text{Select }m\text{ of the }n\text{ boxes.}
\text{There are: }\,{n\choose m}\text{ ways.}$

$\text{Place a ball into each of the }m\text{ boxes.}
\text{Their order does not matter.}$

$\text{Therefore, there are: }\:{n\choose m} \:=\:\frac{n!}{m!\,(n-m)!}$ $\text{ ways.}$

**

 October 21st, 2013, 01:26 AM #3 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Exercise with boxes and balls. The important point to understand is that this problem is the same as choosing m boxes from a total of n boxes. Putting ball into a box equates to "choosing" the box. In the first case, you could imagine the balls being numbered 1-m (but, in fact, the balls could have any distinct numbers or letters or be distinct colours - it doesn't matter). So, you are not only choosing m boxes, but they are ordered. So, this is exactly the same problem as choosing m boxes in order. In the second case, the balls are indistinguishable, so this is the same problem as choosing m boxes where the order does not matter. It's important to see the mathematical/probability equivalence of these problems.
 October 21st, 2013, 03:10 AM #4 Member   Joined: Apr 2013 Posts: 52 Thanks: 0 Re: Exercise with boxes and balls. Nice!!!I got it!!!Thank you very much guys!!!

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