Find the number of strings of length eight of distinct lette

Find the number of strings of length eight of distinct letters of the alphabet so that the words do not have both A and B in them.

First count the number of words that contain both A and B.

8 x 7 x P(24,6)=5,426,910,720

where P(24,6) is permutation,

Therefore the answer is equal to( total number of words of length eight) - ( the number of words of length eight that have both A and B)

then number of strings of eight of distinct letters of the alphabet so that the words do not have both A and B in them=

P(26,8) - 8 × 7 × P(24,6)

P(26,8)= 62,990,928,000

62,990,928,000 - 96,909,120 = 62,894,018,880