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a busy cat so for part a I got the answer as m choose 1 times (1/m)^b

but for part B I am having different approaches and dont know which one is correct

approach 1: m choose 2 times (2/m)^m

approach 2: m choose 2 times (1/m)^m

approach 3: (i was thinking since they are indistinguishable numbers in piles and empty sets)

(n+1-r) choose (r-7)

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2 Answers

Think of the balls, urns as being labelled. That does not affect the probabilities.

There are $m^n$ equally likely ways to distribute the $n$ objects among the $m$ urns.

Call a distribution good if at most $2$ urns contain balls. We count the good distributions.

These are of two types (i) all the balls end up in one urn and (ii) the balls end up in two urns.

It is clear that there are $\binom{m}{1}$ type (i) distributions.

For type (ii), there are $\binom{m}{2}$ ways to choose the two lucky urns. There are $2^n-2$ ways to distribute the balls among the $2$ urns so that neither urn is empty. So the number of distributions of type (ii) is $\binom{m}{2}(2^n-2)$.

Add to find the number of good distributions.

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Thanks. I missed out the -2 part. –  Uma kant May 10 '13 at 15:31
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HINT 2: You have 2 cases (n balls in 1 urn)+(n balls in 2 urns) $$\frac{^mC_1+^mC_2.(2^n-2)}{m^n}$$

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@uma_kant how did u get 2^n - 2 –  jyuserersh May 10 '13 at 18:36
    
@jyuserersh You have to distribute n balls in 2 urns. So each ball has 2 options. Hence $2^n$. The -2 deletes the cases when all the balls go to 1 urn, which we have already calculated. –  Uma kant May 11 '13 at 13:32
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