Five distinct letters are suppose to be transmitted through a communication channel. A total number of $15$ blanks is supposed to be inserted between the first and last letter with at least three between every two. The number of ways in which this can be done is

Place the 5 distinct letters in a row first, which can be done in 5! or 120 ways. This arrangement creates 4 inner gaps between the adjacent letters. The problem requires placing 15 identical blanks into these 4 gaps with at least 3 blanks in each gap. Distribute 3 blanks to each of the 4 gaps first, leaving 15 - 12 = 3 blanks to be freely distributed among the 4 gaps. Using the stars and bars method, the number of ways to distribute the remaining 3 blanks is C(3 + 4 - 1, 4 - 1) = C(6, 3), which equals 20. Multiplying the letter arrangements by the blank distributions gives 120 times 20, which is 2400.

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