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But let's use the standard combinatorial method:
To place \(k = 4\) non-adjacent H’s in \(n = 7\) positions:
The number of ways is \(\binom{n - k + 1}{k} = \binom{7 - 4 + 1}{4} = \binom{4}{4} = 1\)
Yes, confirmed.
So the answer is \(\boxed{1}\)
An anthropologist records 9 distinct cultural practices in a village, each classified as either communal (C) or individual (I). How many sequences of classifications contain at least one pair of consecutive communal practices?
Total sequences: \(2^9 = 512\)
Let \(b_n\) be the number of binary sequences of length \(n\) with no two consecutive C’s (C = communal).
Same recurrence: \(b_n = b_{n-1} + b_{n-2}\)
Base: