Suppose there are MM supporters of Brazil and NN supporters of Argentina. If all of them are made to stand in one line, there will be  M+N1M+N-1 number of places where the fights can occur. Suppose the probability of the fights taking place anywhere is pp.

The chances of fights are similar at each place. So the average of total number of fights is (M+N1)×p(M+N-1)\times p

Now, we have to determine the value of pp. First, let's determine how many ways fights can occur in one place. There must be A,BA,B or B,AB,A for a fight to take place and how many ways other M1M-1 Brazil supporters and N1N-1 Argentina supporters can create new permutations. So the number of fights in one place is 2×(M+N2)!(M1)!×(N1)!2\times \frac{(M+N-2)!}{(M-1)!\times (N-1)!}. Now the average of this value will be pp. That means if we divide it by the total combination, we get pp.

So the formula to determine pp is 2×(M+N2)!(M1)!×(N1)!(M+N)!M!×N!=2×M×N(M+N)×(M+N1)\frac{2\times \frac{(M+N-2)!}{(M-1)!\times (N-1)!}}{\frac{(M+N)!}{M!\times N!}} = \frac{2\times M\times N}{(M+N)\times (M+N-1)}

So the answer is (M+N1)×p=2×M×NM+N(M+N-1)\times p = \frac{2\times M \times N}{M+N}

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