Solving Coin Toss Problem If a coin is tossed 3 times,there are possible 8 outcomes.
HHH HHT HTH HTT THH THT TTH TTT
In the above experiment we see 1 sequnce has 3 consecutive H, 3 sequence has 2 consecutive H and 7 sequence has at least single H.
Suppose a coin is tossed n times.How many sequence we will get which contains a consequnce of H of length at least k?.How to solve this problem using recursive relation?

Thinking about your problem i am sure permutation with repetition would solve it too.
$ \frac{n!}{k!(n-k)!} $ with $k=0,...,n$
