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Given $20$ balls labeled $1$ to $20$, and $10$ bins. How many permutations (order matters) are there if we put $2$ balls in each bins?

My attempt:

Take the first bin, there are ${^{20}\text{P}_2}$ ways of choosing $2$ balls to go into it. Then, there are ${^{18}\text{P}_2}$ ways of choosing. And so on.

So in total, there are $\Pi^{10}_{n=1} \,{^{n}\text{P}_2}$ ways.

Question:

Is this correct?

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I'm assuming you mean $\Pi^{10}_{n=1} \,{^{2n}\text{P}_2}$, right? (I'm assuming $^nP_k = {n \choose k}$). –  celtschk Sep 12 '12 at 17:51
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Assuming you really meant $^{2n}P_2$, there is a big simplification you can do on this, by noting that $\binom{2n}{2} = \frac{2n!}{2(2n-2)!}$ and multiplying out. –  Thomas Andrews Sep 12 '12 at 17:54
    
Doh yes you guys are right. It should be $2n$ instead of $n$ in there. –  Legendre Sep 12 '12 at 18:13
    
Does the order in the bins matter? I would expect not. Your symbol ${}^{m}P_k$ usually stands for permutations so ${}^mP_2=(m)(m-1)$. If order in bins doesn't matter, that is wrong. Maybe you meant ${}^mC_2$. –  André Nicolas Sep 12 '12 at 19:27
    
The order mattered. I used bins and balls for convenience. Perhaps tall slim bins that results in one ball on top of the other? –  Legendre Sep 12 '12 at 20:06

1 Answer 1

up vote 3 down vote accepted

Yes, you are right (apart from the missing factor $2$ I noted in the comment). However it's a quite complex way to write that number.

An easier way is to notice that exchanging the balls in one bin with each other doesn't change anything; since there are $10$ bins, this gives $2^{10}$ such permutations. Since there would be $20!$ permutations otherwise, this means that the total number of permutations is $20!/2^{10}$. Indeed, this can be extended to the case that you have $n$ bins and $kn$ balls, so you put $k$ balls into each bin. Then you'll have $k!$ permutations in each bin which doesn't matter, therefore the total number of different permutations is $(kn)!/(k!)^n$.

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Nice answer! Thanks. –  Legendre Sep 12 '12 at 18:15

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