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November 4th, 2016, 02:08 PM  #1 
Newbie Joined: Oct 2016 From: iran Posts: 1 Thanks: 0  7 digital number combinations can be formed from the {1,2}
How many 7 digital number combinations can be formed from the {1,2} if no some number 1 can appear together? example: 1121222 is not Allowed but 1221212 is allowed. 
November 4th, 2016, 02:44 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 16,368 Thanks: 1172 
34 (a Fibonacci number) if 12 and 21 are considered to be different combinations.

December 26th, 2016, 05:04 AM  #3 
Member Joined: Oct 2016 From: labenon Posts: 33 Thanks: 4 
Let a(k) denote the number of sequences of length k that do not contain two consecutive 1's. Then a1=2 since both the sequences 1 and 2 are permissible and a2=3 since the sequences 12, 21, and 22 are permissible while the sequence 11 is not permitted. Let k≥3. Since a permissible sequence cannot contain consecutive 1's, for a sequence of length k to end in a 1, it must be preceded by a 2. Thus, a permissible sequence of length k that ends in 1 can only be formed by appending the sequence 21 to the end of a permissible sequence of length k−2, of which there are ak−2. A permissible sequence of length k that ends in a 2 can be formed by appending a 2 to the end of a permissible sequence of length k−1, of which there are ak−1. Hence, the number of permissible sequences of length k is given by the recurrence relation a1a2ak=2=3=ak−1+ak−1if k≥3 You can use the recurrence relation to determine a7. 
December 26th, 2016, 05:50 AM  #4 
Senior Member Joined: Feb 2010 Posts: 588 Thanks: 82 
The most number of 1's you can have is four and thus the least number of 2's you can have is three 1212121 > 1 way to do this or C(4,4) ways With four 2's we have _2_2_2_2_ you can fill the blanks with 1's in C(5,3) ways. With five 2's we have _2_2_2_2_2_ and you can fill with 1's in C(6,2) ways. Six 2's ... in C(7,1) ways and seven 2's in C(8,0) ways. So the answer is C(4,4) + C(5,3) + C(6,2) + C(7,1) + C(8,0) = 1 + 10 + 15 + 7 + 1 = 34 ways. This agrees with Skipjack's answer. 

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