Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1986 based on Head Tail Problem.
In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head , a head is immediately followed by ahead and etc. We denote these by TH, HH, and etc. For example in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH, three HT, four TH and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT subsequences?
Integers
Combinatorics
Algebra
Answer: is 560.
AIME I, 1986, Question 13
Elementary Algebra by Hall and Knight
Let us observe the sequences.
H switches to T three times, T switches to H four times.
There are 5 TT subsequences.
We are to add 5 T's into, the string. There are already 4 T's in the sequence.
We are to add 5 balls in 4 urns which is same as 3 dividers \({5+3 \choose 3}\)=56
We do the same with 2H's to get \({2+3 \choose 3}\)=10
so, \(56 \times 10\)=560.
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