Base pairs and sequence

Seven

Please explain the answer sir/mam _/\_

I believe here's how to approach the problem

First, ask yourself, how many ways can you select the first nucleotides: 4

Again, how many ways can you select the second: 4

So far, 4 \(\times\) 4 (to understand why we multiply, refer to the fundamental counting principle in mathematics -- video for reference provided below).

In fact, that's the case for the first 9 nucleotides, so \(4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4\cdot 4=4^9\).

The tenth nucleotide is an M. From the condition provided in the question, there are only 2 ways you can select for M, either A or C.

The next 4 nucleotides, again, are all 4, therefore 44, and the last can be selected in 2 ways again since it is an M.

\(4^9 \cdot 2 \cdot 4^4 \cdot 2 = 4^{14}\)

\((2 \times 10^9) ÷ 4^{14} \approx 7\)