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February 23rd, 2019, 06:58 AM  #11  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 114 Thanks: 2 Math Focus: Calculus  Quote:
Like for the base of that figure $12 \times 9 = 108$ Is this was it was intended to use those numbers?. To get the total by only a multiplication?. This step in the middle is where I'm stuck at. For this reason I need an additional help with what to do next.  
February 23rd, 2019, 07:22 AM  #12  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 114 Thanks: 2 Math Focus: Calculus  Quote:
Why is it a multiplication and not a sum?. I'm not understanding this part very clearly. Mind helping me with this? I tried to understand the use of multiplication, let's say in the second row (where it appears the O's). there are two choices that can be made but looking from the top where do appear the D's, so this would account $4 \times 2 = 8$ so at this step the total is $8$. But on the third, there are $6$ S's so wouldn't meant $5 \times 2 = 10$ to be the total?. By looking again and considering turning left and turning right (zig zag routes), I found that each time you get from the row in the top to the bottom the choices doubles. In the first row there are $4$ ways, in the second row $8$ ways, in the third row $16$ ways, and so on, hence by the time I reach the last step in the figure is $2^{10}=1024$ ways. If I go on with that pattern, this can justify why there is a multiplication. Is this the intended justification Romsek?. Sorry but I'm a slow learner. I'd just hope to be understanding it correctly. There is also a question which arises from all of this. Let's suppose that there were $5$ letters in the top and the number of steps in that pyramid is unchanged, would this meant that the number of ways to read that imaginary word to be $5 \times 2^8$?. Or is it just that such scenario cannot happen because $5$ is not an even number and from this there cannot be possible to have valid two choices?. I just don't want to get confused.  
February 23rd, 2019, 10:12 AM  #13 
Senior Member Joined: Sep 2015 From: USA Posts: 2,590 Thanks: 1434 
I suggest you painstakingly work out a couple of small examples. Perhaps a 3 level pyramid first with only one starting point, then with two. Label each point in the pyramid carefully and see how many sequences you can come up with that spell your 3 letter word. That will teach you what's going on much better than reading my explanation of it will. 
February 23rd, 2019, 10:59 AM  #14 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 
There are 4 choices for the initial D. We can then go right or left. So for the first two letters we have 2 * 4 = 4 * 2^1 choices. For the third letter, we can again go left or right. So for the first 3 letters we have 2 * (4 * 2^1) = 4 * 2^2 choices. Generalizing, we have 4 * 2^(n  1) choices for the first n letters. The phrase has 9 letters so the answer is 4 * 2^(9  1) = 2^2 * 2^8 = 2^(10) = 1024 
February 24th, 2019, 11:57 AM  #15  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 114 Thanks: 2 Math Focus: Calculus  Quote:
It took some time but I reanalyzed my drawing and found this from following the possible routes. There are $256$ for each D so in total $4$ times that amount, hence $1024$. Now there was an unattended question. What if we're facing a 5 letter word in the top and not a four letter word. Then the answer would follow the same "recipe" hence it would be $5 \times 2^{8}$. This whole thing it looks to be a way to play with combinatorics.  
February 24th, 2019, 12:00 PM  #16  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 114 Thanks: 2 Math Focus: Calculus  Quote:
 
February 24th, 2019, 01:57 PM  #17 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
Jeff, you forgot to include a sketch...whassamattayou?!


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