Maximum Parsimony with nucleotide data

Informative SNP positions of types 5, 6, & 7 favor the first, second, or third hypotheses of relationship, respectively, because each of these types requires a single nucleotide change in that hypothesis, and two changes in the two alternative hypotheses.They are informative because they provide evidence in favor of one & only one hypothesis over the others.

Then, in a hypothetical data set [not shown] with 10 Type 5 positions, 5 Type 6 positions, and 2 Type 7 positions, 
    Hypothesis #1 then requires  N = 10 + 2(  5 + 2) = 24 changes
    Hypothesis #2 would require N =   5 + 2(10 + 2) = 29 changes
    Hypothesis #3 would require N =   2 + 2(10 + 5) = 32 changes

Thus, the first hypothesis requires the fewest changes and is therefore the most parsimonious explanation of the data. Note that this hypothesis has the largest number of parsimony sites that favor it.

Uninformative SNP positions of types 1, 2, 3, & 4 uniformly require 0, 1, 3, & 2 nucleotide changes, respectively, in any of the hypotheses. They are uninformative because they do not provide evidence in favor of any one of the three hypotheses over the others. [SNP positions of types 3 & 4 may be informative, under certain models of molecular evolution].

HOMEWORK: A students asks, "The most parsimonious explanation has the largest number of sites that favor it. I don't understand: that seems backwards." Explain the student's error.

HOMEWORK: SHOW that positions of types 1, 2, 3, & 4 require 0, 1, 3, & 2 nucleotide changes, respectively, for any of the three hypotheses above. Hint: Draw the three hypotheses, and find at least one assignment of changes to the branches or internode that satisfies the statement.