Let mn distinct coins have been distributed into m different purses, n coins into each purse. If p is probability that two specified coins will be found in the same purse, then

Given that  mn coins have been distributed into  na purses, n into each
        Therefore, we have  $\frac{mn}{na}=\frac{m}{a}=n$
       Total number of purses is  na and the total number of coins is  mn
       Now there are two specified coins. Let it be A and B .
      A Purse contains n coins. So if A is placed into one purse, there remains n-1 positions. At the same time, on the whole there remains mn-1 coins.