My Math Forum How to find the different number and ways to read a given phrase forming a pile?

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February 23rd, 2019, 06:58 AM   #11
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Quote:
 Originally Posted by Denis Wasn't mad! 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
Okay!. I did understood that those letters referred to the numbers but my question was what to do with those numbers?. Should I sum them?

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
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Quote:
 Originally Posted by romsek step 1 2 choices step 2 2 choices each step 2 choices 2 x 2 x 2 x ... 2 8 times = 2^8

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: 2,452 Thanks: 1337 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: 551 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
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Quote:
 Originally Posted by romsek 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.
Well that's what I did, and I found that each time you go in the second step the choices doubles and so forth. Probably this is the reason why it is a power of two. I've also noticed that there are four choices by looking from the D's in the top. In the end it is like tossing a coin, each new coin is added is another set of two choices, hence increasing by two times the total.

It took some time but I re-analyzed 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
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Quote:
 Originally Posted by JeffM1 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
Thanks! This explanation added to my findings explained in an earlier post really gave me an idea of what was happening. Although it would had been much more helpful to get a sketch to accompany those words.

 February 24th, 2019, 01:57 PM #17 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038 Jeff, you forgot to include a sketch...whassamattayou?!

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