<PRE><B>Question 1150912</B>
Hope this helps you reason out the answer:&lt;br&gt;

if 2n=2:  HT and TH  are the only ways to get 1 head --&gt; n=1 so this is n+1 ways &lt;br&gt;

if 2n=4:   HTHT, HTTH, and THTH   --&gt;  3 ways to get 2 nonadjacent heads --&gt;  and since n=2, this is also n+1 ways&lt;br&gt;

(a) if 2n=6:  HTHTHT, HTTHTH, HTHTTH, and THTHTH  --&gt; &lt;b&gt;4 ways&lt;/b&gt; to get 3 nonadjacent heads, n=3 so this is yet again n+1 ways &lt;br&gt; 

(b) This implies that in general for 2n tosses, there are ____ ways to get n nonadjacent heads&lt;br&gt;
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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&#39;s:  1+1+(n-1) = n+1

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