Consider the 3 boys as a single unit or 'block'. This reduces the problem to arranging 10 units: 1 block of boys, 4 girls, and 5 adults.

Since an adult must be at each end, we choose 2 adults to be at the ends. The number of ways to choose 2 adults from 5 is given by:

\(^5P_2 = 5 \times 4 = 20\)

Now, arrange the remaining 8 units (1 block of boys, 4 girls, and 3 adults) in the middle:

\(8! = 40,320\)

Within the block of boys, the 3 boys can be arranged among themselves in:

\(3! = 6\)

Therefore, the total number of arrangements is:

\(8! \times 3! \times ^5P_2 = 40,320 \times 6 \times 20 = 4,838,400\)