I want someone to check if I'm right or not. The problem seems to be easy but everybody knows that not always effortless problems are straightforwardly solvable.
Given symbols: A, B, C and D and create a 39 symbols sequence out of those four symbols. What's the probability that the 4 A's are in the beggining and in the sequence are 37 A's at all?
My solution: I believe the answer should be $\frac{ 3 \cdot\left( \begin{array}{cc} 35 \\ 33,1,1,0 \end{array}\right) + 3 \cdot \left( \begin{array}{cc} 35 \\ 33,2,0,0 \end{array}\right) }{4^{39}}$. There are $4^{39}$ combinations at all. . I know it doesn't look nice. Can someone check it or suggest other solution?