. Using Stirling’s approximation, we found an approximation for #A, the number of alignments. Apply that formula to n = m = 1000. Why didn’t we simply compute _ 2000 1000 _ exactly on our computer?  ...

.
Using Stirling’s approximation, we found an approximation for

#A, the number of alignments. Apply that formula to n = m = 1000. Why

didn’t we simply compute _

2000

1000

_

exactly on our computer?

.
If we use a gap penalty g(k) = −10k, we observe that all optimal

alignments of strings A and B are identical for all values of the mismatch

penalty μ when μ > 20. Explain why this is true.