Question 1150912: Find the number of ways a coin can be tossed:
(a) 6 times so that there is exactly 3 heads and no two heads occur in a row.
(b) 2n times so that there is exactly n heads and no two heads occur in a row.
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! Hope this helps you reason out the answer:
if 2n=2: HT and TH are the only ways to get 1 head --> n=1 so this is n+1 ways
if 2n=4: HTHT, HTTH, and THTH --> 3 ways to get 2 nonadjacent heads --> and since n=2, this is also n+1 ways
(a) if 2n=6: HTHTHT, HTTHTH, HTHTTH, and THTHTH --> 4 ways to get 3 nonadjacent heads, n=3 so this is yet again n+1 ways
(b) This implies that in general for 2n tosses, there are ____ ways to get n nonadjacent heads
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There is one way to toss HTHTHT...HT, and one way to toss THTHTH...TH and for cases HT....TH (starting and ending with H) there are always n-1 places to put TT between two of the H's: 1+1+(n-1) = n+1
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