
Applied Math Applied Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 16,225 Thanks: 1150 
34 (a Fibonacci number) if 12 and 21 are considered to be different combinations.

December 26th, 2016, 05:04 AM  #3 
Newbie Joined: Oct 2016 From: labenon Posts: 28 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: 583 Thanks: 80 
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. 

Tags 
combinations, digital, formed, number 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Number of combinations?  Apple30  Elementary Math  13  August 26th, 2016 11:45 PM 
Max number of combinations  Piggy  Elementary Math  4  July 8th, 2016 03:42 PM 
prove that angle formed at orthocenter is supplement of the angle formed at the vertx  randomgamernerd  Geometry  2  November 17th, 2015 09:03 AM 
Number of Combinations?  Pal  Algebra  2  July 15th, 2013 06:55 PM 
Number of closed paths formed by arcs of one fifth of a circ  rhseiki  Number Theory  1  May 17th, 2011 07:44 AM 