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`(.^(N)C_(M)xxN!)/(M^(N))``(.^(N)C_(M)xxM!)/(N^(M))``1-(.^(N)C_(M)xxM!)/(M^(N))``1-(.^(N)C_(M)xxN!)/(N^(M))`

Answer :

BSolution :

Since lines are more than telegrams, `.^(N)C_(M)` are those lines where telegrams will be sent. <br> So, number of favourable cases = `.^(N)C_(M) xx M!` <br> Total number of cases = `N^(M)` <br> [As first telegram can go in any one of n lines] <br> So, required probability = `(.^(N)C_(M)xxM!)/(N^(M))`**Introduction and definition**

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