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If $mn$ coins have been distributed into $m$ purses, $n$ into each find:

$(1)$ The chance that two specified coins will be found in the same purse, and

$(2)$ What the chance becomes when $r$ purses have been examined and found not to contain either of the specified coins

Source:A First Course in Probability by Tapas K. Chandra, Dipak Chatterjee,Page-76,Chapter 1

MY ATTEMPT:

$(1)$ Total number of ways in which $mn$ coins can be distributed in $m$ purses with $n$ coins in each is:

$$\dfrac{(mn)!}{(n!)^mm!}$$ Suppose we have distributed any $n$ coins (containing the specified $2$ coins) in any selected purse, then the probability is: $$\dfrac{\binom{mn}{n-2}\binom{m}{1}\dfrac{(mn-n)!}{(n!)^{m-1}(m-1)!}}{\dfrac{(mn)!}{(n!)^mm!}}$$ However, the answer to the first question does not match. Where am I going wrong?

Also, please give some hints for the second part of the question. Thank you.

1 Answers 1

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(1) Well, one of the coins will certainly be in one of the $m$ purses; but who cares which.   Wheresoever that coin is, then there are $(nm{-}1)$ 'places' the other coin could be, of which $(n{-}1)$ are in the same purse as the first.   There is no need to involve factorials.

(2) Well, then we've just eliminated $~nr~$ 'places' the second coin could be, is all.