Enclave radio problem

1. Note that the conditions means each speech precedes a music piece. (Two speeches cannot be consecutive, and the last one cannot be a speech.) Choose the four music pieces that will be preceded by a speech. One of them must be the first one, so we're picking three out of six: $\binom{6}{3} = 20$ ways.
2. Arrange the four speeches: $4! = 24$ ways.
3. Arrange the seven music pieces: $7! = 5040$ ways.

It's clear that all of them give a valid schedule, and any schedule can be generated in exactly one way. Thus the total number of schedules is just the product of them all: $20 \cdot 24 \cdot 5040 = \boxed{2419200}$ .

Solution:

Firstly, to get the answer, dear Readers, we should find the number of different formats. Now, You may be thinking, "What is format?". Example 1 and 2 have the same format, $|**|**|**|*$ (where $|$ represents any speech and $*$ represents any music piece). Example 3 has the following format: $|****|*|*|*$ .

So we will analyze that. One speech must go on the start and one m.p. must go on end, leaving 3 speeches and 6 m.p. We will "combine" one speech and one m.p. together in the following format: *| that way, there will be 3 music-piece-speech blocks and 3 music pieces left for combining. That ends up to be the standard stars and bars problem. So number of different formats is $\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) = 20$ .

Now, in each of those formats, we can spread music pieces and speeches in the same manner. Speeches can be spread on $4! = 24$ different ways, while music pieces can be on $7! = 5040$ . All that, dear Readers, leaves the final solution to be $\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \times 4! \times 7! = \boxed{2,419,200}$