. Another approach to counting alignments is from the recursion formula B(n,m) = B(n−1,m)+B(n,m−1)+B(n−1,m−1), where B(n,m) is the number of alignments of a1a2 · · · an with b1b2 · · · bm. Using B(0,...

.
Another approach to counting alignments is from the recursion

formula B(n,m) = B(n−1,m)+B(n,m−1)+B(n−1,m−1), where B(n,m)

is the number of alignments of a1a2 · · · an with b1b2 · · · bm. Using B(0, 0) = 0

and B(i, 0) = 1 = B(0, j) (all i and j), use a matrix to find B(5, 5). Compare

this number with

Why is B(5,5) larger?