February 21st, 2019, 08:54 AM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  How to find the different number and ways to read a given phrase forming a pile?
I'm totally lost in this riddle, does it exist a way to calculate the different ways to read a word using a systematically approach?. In my initial attempt what I tried to do is drawing a circle over each time I could identify the word being asked but in the end I got very confused and I felt that I counted twice the word, hence I couldn't even understand if my attempt was right. The problem is as follows: At a kindergarten's playroom in Taichung a teacher assembled the following configuration using alphabet cubes forming a stack (see the figure as a reference) where it can be read the word DOS BANDOS. (Spanish word for two sides). Compute the number of different ways and joining neighboring letters can be read the phrase DOS BANDOS. The given alternatives in my book are: $\begin{array}{ll} 1.&1536\\ 2.&1280\\ 3.&256\\ 4.&768\\ 5.&1024\\ \end{array}$ So as mentioned, from what I could identify immediately was seen from the top to bottom there are four cubes which form vertically the word being asked. From left to right, and then from right to left another two. This accounts for six. My findings are pictured in the diagram from below, colored with orange. But that's how far I went. As the more I looked at the stack I started to get confused on which can be allowed ways and which do already counted. Hence can somebody help me with this riddle? An answer which would help me the most is a way to methodically to solve this rather than just drawing circles over words as if it were a word search puzzle on a newspaper. Overall does it exist a way?. Can somebody help me to be on the right path on this one?. If a diagram or drawing is necessary for an explanation or a justification of the method please include in your answer because I believe an answer for this question would be greatly improved with a visual aid, as I am not good at understanding just plain words or straightforward formulas Please be as vocal as possible. I've been told as a hint that this kind of problem requires considering turns left and right but I don't know how to account for them or how to use this information properly hence a drawing can be necessary for explanation. 
February 21st, 2019, 10:00 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025 
Have you tried it this way: Code: 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 
February 21st, 2019, 10:22 AM  #3 
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  I'm sorry but what did you meant by using this figure?. I don't get the idea. Perhaps you could explain what to do with those numbers?. Where do they appear?. A justification and clarification is needed there. 
February 21st, 2019, 10:49 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025 
Huh? I've replaced the letters with digits. Makes it easier to work with. If you can't see that, then just forget it ! 
February 21st, 2019, 11:28 AM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 
Consider for a moment a pyramid of letters with only a single D at the top rather than 4. As you are working your way from top to bottom you have 2 choices at each step which letter of the next level to choose. You have 8 steps which yields $2^8=256$ total choices to spell out DOSBANDOS With the given truncated pyramid with 4 D's on top we simply have 4 starting points with which to do the described procedure. Thus we have $2^8 \cdot 4 = 1024$ possible paths through neighbors that spell out DOSBANDOS. Choice 5 
February 22nd, 2019, 11:56 AM  #6  
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  Quote:
In other words how does replacing the letters with numbers eases the work or the solution?. It's been days and i'm still intriguing about your hint because I'd like to learn how would it help to solve this riddle. If you accompany an example or develop some steps maybe I can catch up from there and solve it on my own. Mind taking a chance on teaching me?  
February 22nd, 2019, 12:05 PM  #7  
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  Quote:
Basically the part where I'm stuck at now is finding the reason for the usage of $2^8$ as the possible paths. If the number of allowed paths goes from taking right or left, wouldn't it be just enough as I mentioned $8 \times 2$?. Maybe it is a silly question but I'd like to understand what's going on.  
February 22nd, 2019, 01:18 PM  #8  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025  Quote:
Using numbers simply makes it easier to "do the work". D=1, O=2, S=3, B=4, A=5, N=6, D=7, O=8, S=9  
February 22nd, 2019, 01:23 PM  #9 
Senior Member Joined: Dec 2015 From: somewhere Posts: 514 Thanks: 80 
Using numbers you avoid distractions.
Last edited by idontknow; February 22nd, 2019 at 01:26 PM. 
February 22nd, 2019, 01:25 PM  #10 
Senior Member Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 
step 1 2 choices step 2 2 choices each step 2 choices 2 x 2 x 2 x ... 2 8 times = 2^8 

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