Permutations

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Question

The number of ways in which we can get at least 6 successive tails when a coin is tossed 10 times in a row, is

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Solution

The sequence of six tails may begin at with the ith toss where $i = 1, 2, 3, 4, 5 .$ Thus, the number of ways is
$2^{4} + 1 (2^{3}) + 2 (1) (2^{2}) + 2^{2} (1) 2 + 2^{3} (1) = 48$

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