Question

Assume strings contain ONLY the letters: a, b.

How many bit strings of length 8 either start with bbb or end with aa ?

Answer 1

n(start with bbb) = n(fix first 3 as bbb, combinations for remaining 5 letters)

= (no. of option for each letter)^5

= 2^5

= 32

n(end with aa) = n(fix last 2 as aa, combinations for remaining 6 letters)

= (no. of option for each letter)^6

= 2^6

= 64

n(start with bbb and end with aa)

= n(fix last 2 as aa, first 3 as bbb, combinations for remaining 3 letters)

= (no. of option for each letter)^3

= 2^3

= 8

n(start with bbb or end with aa) = n(start with bbb) + n(end with aa) - n(start with bbb and end with aa)

{subtracting n(start with bbb and end with aa) since it is counted twice as it is common in both cases}

= 32+64-8

= 88 (answer)

(please UPVOTE)