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kindly provide the problem 27 from section 1.4 of your text. am waiting to help you answer this question
The textbook isn't needed for this. The rest of the question just stands on it's own.
So, when the log ends with n(S) = 2, n(F) = 0 and n(T) = 0, 1, ..., 4
With n(T) = 0, there is just one possible log: S, S.
With n(T) = 1, there are two possible logs: T, S, S and S, T, S.
With n(T) = 2, the number of logs is the number of ways of choosing which position 1, 2 or 3 the S will come, which is 3,
With n(T) = 3, the number of logs is 4.
With n(T) = 4, the number of logs is 5.
Total for n(S) = 2 is 1 + 2 + 3 + 4 + 5 = 15
If n(T) = 5, then once again n(F) = 0, and n(S) = 0 or 1.
If n(S) = 0, there is just one log.
If n(S) = 1, there are 5 positions in which the S could come.
So the total for n(T) = 5 is 1 + 5 = 6
If n(F) = 1, then the F must come at the end of the sequence, with n(S) = 0, 1 and n(T) = 0, 1, ..., 4
If n(S) = 0, there are 5 possible logs, corresponding to each of the five values of n(T).
If n(S) = 1, there are 1 + 2 + 3 + 4 + 5 = 15 possible logs (each of
these individual numbers being the number of ways of positioning the S
with 0, 1, ..., 4 T's)
So the total for n(F) = 1 is 5 + 15 = 20
So I reckon the overall total is 15 + 6 + 20 = 41 logs
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