Related Articles
- Case 2: The three close pairs form a path: \( AB, BC, CD \). Edges: AB, BC, CD. Trio \( A,B,C \): AB, BC close â 2 close pairs. Trio \( B,C,D \): BC, CD close â 2 close pairs. Trio \( A,B,D \): AB, BD? BD not close, AD not close â only AB â 1 close â fails. Trio \( A,C,D \): AC not close â 0 â fails. So only the two middle trios (A,B,C) and (B,C,D) each have exactly 2 close pairs.
- So one two-edge path with distinct middle vertex.
- Number of such paths: a path of 3 nodes has 2 endpoints and one middle. Number of such paths in a complete graph of 5 nodes: number of ways to choose 3 distinct nodes: \( inom{5}{3} = 10 \), each induces a path unless it's complete or linear.
- Alternate approach: total number of trios with exactly two close pairs.
- Each such trio corresponds to a 2-edge path (i.e., two adjacent edges sharing a vertex, not forming a triangle).
- For each internal node of degree at least 2 in the close pair graph, we can count how many such paths go through it.
Trending Articles
Shocking Minecraft Pocket Edition Mod Revealed – Transform Your Experience NOW!
A virologist is studying the combinations of synthetic DNA sequences where two types of nucleotides, A and T, are used. How many 6-nucleotide sequences can be formed where there are exactly two consecutive A's at least once?
To solve this, we first compute the total number of 6-nucleotide sequences using A and T, which is \(2^6 = 64\).
Next, we compute the number of sequences that do **not** contain two consecutive A’s. Let’s denote this count as \(a_n\) for sequences of length \(n\).
We derive a recurrence for \(a_n\):
If the first nucleotide is T, the rest \(n-1\) positions form a valid sequence: \(a_{n-1}\)
If the first is A, the next must be T (to avoid AA), and then \(n-2\) positions follow: \(a_{n-2}\)
So, \(a_n = a_{n-1} + a_{n-2}\), with base cases:
\(a_1 = 2\) (A, T)
\(a_2 = 3\) (TT, TA, AT — not AA)
