The first subtask can be solved by iterating over all possible subsets and counting the number of similar pairs in each one.

Since each element is either in the chosen subset or not, that leads us to a dynamic programming solution, where ~dp[n][k]~ is the number of ways to choose among the first ~n~ words a subset with exactly ~k~ similar pairs. The transition would be:

$$\displaystyle dp[n][k] = dp[n-1][k]+dp[n-1][k-x]$$

where ~x~ is the number of new similar pairs that appear when we add the ~n^\text{th}~ word to the subset. Unfortunately, since we don't know which words are in the subset, we can't know ~x~.

But, we can use a similar approach if we notice that if we sort the letters in each word, then the words are similar if and only if they are equal.

After sorting, we get a multiset of words, with ~m~ distinct elements, and the ~i^\text{th}~ element appears ~a_i~ times. Let ~dp[m][k]~ be the number of ways to choose among the first ~m~ different words a subset with exactly ~k~ equal words. The transition is:

$$\displaystyle dp[m][k] = \sum_{i=0}^{i \le a_i, \frac{i(i-1)}{2} \le k} \binom{a_i}{i} dp[m-1][k-i(i-1)/2]$$

The initial values are ~dp[0][k] = 1~, for ~k = 0~, or ~0~ otherwise.

Since the numbers in the binomial coefficient are at most ~n~, we can precompute them using Pascal's triangle. The complexity for each ~k~ is ~\mathcal O(n)~, since the sum of all ~a_i~ is equal to ~n~, so the total complexity of the solution is ~\mathcal O(n^2+nk)~.