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Is it possible to methodically find the total of ways to read a given phrase making a stack?


Sorting cards using max 3 stacksAll Words Go to Heaven?Find the number given its remaindersThe poorly designed clockTwo IQ Test puzzles nobody knows an answer toWhat are the missing patterns? (From a 5th grader's school book)Lets see how smart Puzzling' stack users really are!How are the words weighed?This is a Mensa IQ test from Norway. The last puzzle.Obscure IQ test question













1












$begingroup$


I have been going in circles for a few days now with this word-search puzzle and I still cannot find a clever solution to the problem. Is there a systematic approach that doesn't involve guessing? 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 (the Spanish word for two sides). Calculate the number of different ways joining neighboring letters can be read the phrase DOS BANDOS.




sketch of the problem



The possible solutions in my book are:



$begin{array}{ll}
1.&1536\
2.&1280\
3.&256\
4.&768\
5.&1024\
end{array}$



In my initial attempt what I tried to do is to draw 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 a possibility twice, hence I couldn't even understand if my attempt was right. During this process I could identify immediately that the word could be read from the top to bottom, there I counted four cubes, going from left to right, hence $4$ ways and from right to the left $2$ ways which together would account for $6$. This is summarized in the drawing from below where the circles are painted with blue color.



Sketch of the solution



In the end that how far I went. As the more I looked at the stack I started to get confused on which zig zag lines are allowed and which do are already counted.



Therefore can somebody help me with this riddle?. To be honest I have little experience with these kinds of problems so I'd like somebody could be as much as detailed possible and include some drawing (perhaps using mine as a reference) and justify a method for solution.



It is very important for me to get a visual aid, because I really don't feel that solely a paragraph alone would be enough to understand, even as hard as I could. So really please if you can help me with this, include some sort of drawing or schematic so I can understand how to calculate the number of ways.



Again, I am looking for an answer which can solve this problem and that it can be extended to similar problems. Does it exist a way? I've been told a hint which mentions that I should consider turns to the left or right. But I don't know how to use this information.



There is also, in the bottom of the truncated pyramid it can be read the word DOS. Would this count if going from bottom to the top?.



Overall I hope somebody can take time and answer these questions. Because I really need help with this one and I'm confused.










share|improve this question











$endgroup$








  • 1




    $begingroup$
    Hey! This is a lovely question that has been answered by both ppgdev and myself below, but really belongs on Math.SE. I've flagged the question as such.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    I disagree. It's such a famous puzzle that we should have it here
    $endgroup$
    – Dr Xorile
    2 hours ago










  • $begingroup$
    @Dr Xorile I didn't know that this was a famous puzzle?. Do you know it's origin?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago
















1












$begingroup$


I have been going in circles for a few days now with this word-search puzzle and I still cannot find a clever solution to the problem. Is there a systematic approach that doesn't involve guessing? 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 (the Spanish word for two sides). Calculate the number of different ways joining neighboring letters can be read the phrase DOS BANDOS.




sketch of the problem



The possible solutions in my book are:



$begin{array}{ll}
1.&1536\
2.&1280\
3.&256\
4.&768\
5.&1024\
end{array}$



In my initial attempt what I tried to do is to draw 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 a possibility twice, hence I couldn't even understand if my attempt was right. During this process I could identify immediately that the word could be read from the top to bottom, there I counted four cubes, going from left to right, hence $4$ ways and from right to the left $2$ ways which together would account for $6$. This is summarized in the drawing from below where the circles are painted with blue color.



Sketch of the solution



In the end that how far I went. As the more I looked at the stack I started to get confused on which zig zag lines are allowed and which do are already counted.



Therefore can somebody help me with this riddle?. To be honest I have little experience with these kinds of problems so I'd like somebody could be as much as detailed possible and include some drawing (perhaps using mine as a reference) and justify a method for solution.



It is very important for me to get a visual aid, because I really don't feel that solely a paragraph alone would be enough to understand, even as hard as I could. So really please if you can help me with this, include some sort of drawing or schematic so I can understand how to calculate the number of ways.



Again, I am looking for an answer which can solve this problem and that it can be extended to similar problems. Does it exist a way? I've been told a hint which mentions that I should consider turns to the left or right. But I don't know how to use this information.



There is also, in the bottom of the truncated pyramid it can be read the word DOS. Would this count if going from bottom to the top?.



Overall I hope somebody can take time and answer these questions. Because I really need help with this one and I'm confused.










share|improve this question











$endgroup$








  • 1




    $begingroup$
    Hey! This is a lovely question that has been answered by both ppgdev and myself below, but really belongs on Math.SE. I've flagged the question as such.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    I disagree. It's such a famous puzzle that we should have it here
    $endgroup$
    – Dr Xorile
    2 hours ago










  • $begingroup$
    @Dr Xorile I didn't know that this was a famous puzzle?. Do you know it's origin?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago














1












1








1


1



$begingroup$


I have been going in circles for a few days now with this word-search puzzle and I still cannot find a clever solution to the problem. Is there a systematic approach that doesn't involve guessing? 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 (the Spanish word for two sides). Calculate the number of different ways joining neighboring letters can be read the phrase DOS BANDOS.




sketch of the problem



The possible solutions in my book are:



$begin{array}{ll}
1.&1536\
2.&1280\
3.&256\
4.&768\
5.&1024\
end{array}$



In my initial attempt what I tried to do is to draw 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 a possibility twice, hence I couldn't even understand if my attempt was right. During this process I could identify immediately that the word could be read from the top to bottom, there I counted four cubes, going from left to right, hence $4$ ways and from right to the left $2$ ways which together would account for $6$. This is summarized in the drawing from below where the circles are painted with blue color.



Sketch of the solution



In the end that how far I went. As the more I looked at the stack I started to get confused on which zig zag lines are allowed and which do are already counted.



Therefore can somebody help me with this riddle?. To be honest I have little experience with these kinds of problems so I'd like somebody could be as much as detailed possible and include some drawing (perhaps using mine as a reference) and justify a method for solution.



It is very important for me to get a visual aid, because I really don't feel that solely a paragraph alone would be enough to understand, even as hard as I could. So really please if you can help me with this, include some sort of drawing or schematic so I can understand how to calculate the number of ways.



Again, I am looking for an answer which can solve this problem and that it can be extended to similar problems. Does it exist a way? I've been told a hint which mentions that I should consider turns to the left or right. But I don't know how to use this information.



There is also, in the bottom of the truncated pyramid it can be read the word DOS. Would this count if going from bottom to the top?.



Overall I hope somebody can take time and answer these questions. Because I really need help with this one and I'm confused.










share|improve this question











$endgroup$




I have been going in circles for a few days now with this word-search puzzle and I still cannot find a clever solution to the problem. Is there a systematic approach that doesn't involve guessing? 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 (the Spanish word for two sides). Calculate the number of different ways joining neighboring letters can be read the phrase DOS BANDOS.




sketch of the problem



The possible solutions in my book are:



$begin{array}{ll}
1.&1536\
2.&1280\
3.&256\
4.&768\
5.&1024\
end{array}$



In my initial attempt what I tried to do is to draw 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 a possibility twice, hence I couldn't even understand if my attempt was right. During this process I could identify immediately that the word could be read from the top to bottom, there I counted four cubes, going from left to right, hence $4$ ways and from right to the left $2$ ways which together would account for $6$. This is summarized in the drawing from below where the circles are painted with blue color.



Sketch of the solution



In the end that how far I went. As the more I looked at the stack I started to get confused on which zig zag lines are allowed and which do are already counted.



Therefore can somebody help me with this riddle?. To be honest I have little experience with these kinds of problems so I'd like somebody could be as much as detailed possible and include some drawing (perhaps using mine as a reference) and justify a method for solution.



It is very important for me to get a visual aid, because I really don't feel that solely a paragraph alone would be enough to understand, even as hard as I could. So really please if you can help me with this, include some sort of drawing or schematic so I can understand how to calculate the number of ways.



Again, I am looking for an answer which can solve this problem and that it can be extended to similar problems. Does it exist a way? I've been told a hint which mentions that I should consider turns to the left or right. But I don't know how to use this information.



There is also, in the bottom of the truncated pyramid it can be read the word DOS. Would this count if going from bottom to the top?.



Overall I hope somebody can take time and answer these questions. Because I really need help with this one and I'm confused.







pattern visual calculation-puzzle






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 hours ago









Hugh

2,0591923




2,0591923










asked 3 hours ago









Chris Steinbeck BellChris Steinbeck Bell

1163




1163








  • 1




    $begingroup$
    Hey! This is a lovely question that has been answered by both ppgdev and myself below, but really belongs on Math.SE. I've flagged the question as such.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    I disagree. It's such a famous puzzle that we should have it here
    $endgroup$
    – Dr Xorile
    2 hours ago










  • $begingroup$
    @Dr Xorile I didn't know that this was a famous puzzle?. Do you know it's origin?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago














  • 1




    $begingroup$
    Hey! This is a lovely question that has been answered by both ppgdev and myself below, but really belongs on Math.SE. I've flagged the question as such.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    I disagree. It's such a famous puzzle that we should have it here
    $endgroup$
    – Dr Xorile
    2 hours ago










  • $begingroup$
    @Dr Xorile I didn't know that this was a famous puzzle?. Do you know it's origin?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago








1




1




$begingroup$
Hey! This is a lovely question that has been answered by both ppgdev and myself below, but really belongs on Math.SE. I've flagged the question as such.
$endgroup$
– Hugh
2 hours ago




$begingroup$
Hey! This is a lovely question that has been answered by both ppgdev and myself below, but really belongs on Math.SE. I've flagged the question as such.
$endgroup$
– Hugh
2 hours ago












$begingroup$
I disagree. It's such a famous puzzle that we should have it here
$endgroup$
– Dr Xorile
2 hours ago




$begingroup$
I disagree. It's such a famous puzzle that we should have it here
$endgroup$
– Dr Xorile
2 hours ago












$begingroup$
@Dr Xorile I didn't know that this was a famous puzzle?. Do you know it's origin?.
$endgroup$
– Chris Steinbeck Bell
2 hours ago




$begingroup$
@Dr Xorile I didn't know that this was a famous puzzle?. Do you know it's origin?.
$endgroup$
– Chris Steinbeck Bell
2 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$


The answer is - the number of different ways joining neighboring letters to get the phrase DOS BANDOS is 1024.


You can record your path down as a sequence of 1s and 0s - left turn 0, right turn 1. Your record from any of the 4 D's at the top could be any combination of 8 ones/zeros. It gives you a total of 256 paths. Multiply it by 4 possible Ds to start with and you get the answer. Below is an example of a path record:




                D
/ left 0
O O
right 1
S S S
right 1
B B B B
/ left 0
A A A A A
right 1
N N N N N N
right 1
D D D D D D D
/ left 0
O O O O O O O O
/ left 0
S S S S S S S S S





share|improve this answer










New contributor




ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Can you include a drawing in your answer so I could understand what were you referring with the $1$ and $0$ ?. Because I'm not getting the idea very clear. I believe there are zig zag options but I don't know how to account for them. Can you help with this part please?. :)
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell, I added an example of a recorded path, as you asked. I noticed Hugh has already helped you with understanding the answer. I hope the picture will also help clarifying it.
    $endgroup$
    – ppgdev
    1 hour ago



















1












$begingroup$

The solution to this problem is...




$1024$


From any given circle that is not in the bottom row, you can either go down and to the left or down and to the right. Because there are four starting points and eight rows in which we make a turn, the solution is $4 times 2^{8} = 1024$.







share|improve this answer











$endgroup$













  • $begingroup$
    Precisely this is what I wanted to avoid. Can you explain why is it a power of $2$?. In my question what I felt was needed for me to understand the rationale behind this question was to accompany some drawing to justify an answer. Because I'm still stuck on how to get the answer.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell think about it this way: say you have one coin; it can be either "heads" or "tails" — that's $2$ possibilities. Now, add a second coin; now you can have "heads-heads", "heads-tails", "tails-heads", or "tails-tails" — that's $4$ possibilities. Now, add a third coin; now, you can have "heads-heads-heads", "heads-heads-tails", "heads-tails-heads", .... etc. for a total of $8$ possible arrangements. Notice that each time we add a new coin, the number of possibilities doubles [cont'd]
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    that's a more logical way!. But why each time a coin is added the result is multiplied and not a sum?. I mean why the possibilities do not just sum?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell [cont'd] this is because the new coin introduces another combination. Let me do an example. We already know that $3$ coins = $8$ possibilities, so let's try and mathematically calculate the number of combinations for $4$ coins. For each of the $8$ combinations with $3$ coins, we can add a "heads" to the end, or a "tails" to the end. That means that the number of possibilities doubles.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    To reference your comment as to why we multiply and not add, let me put it this way. Say that we're trying to calculate the number of possibilities for $10$ coins. Now, take a look at this picture, or this picture
    $endgroup$
    – Hugh
    2 hours ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$


The answer is - the number of different ways joining neighboring letters to get the phrase DOS BANDOS is 1024.


You can record your path down as a sequence of 1s and 0s - left turn 0, right turn 1. Your record from any of the 4 D's at the top could be any combination of 8 ones/zeros. It gives you a total of 256 paths. Multiply it by 4 possible Ds to start with and you get the answer. Below is an example of a path record:




                D
/ left 0
O O
right 1
S S S
right 1
B B B B
/ left 0
A A A A A
right 1
N N N N N N
right 1
D D D D D D D
/ left 0
O O O O O O O O
/ left 0
S S S S S S S S S





share|improve this answer










New contributor




ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Can you include a drawing in your answer so I could understand what were you referring with the $1$ and $0$ ?. Because I'm not getting the idea very clear. I believe there are zig zag options but I don't know how to account for them. Can you help with this part please?. :)
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell, I added an example of a recorded path, as you asked. I noticed Hugh has already helped you with understanding the answer. I hope the picture will also help clarifying it.
    $endgroup$
    – ppgdev
    1 hour ago
















2












$begingroup$


The answer is - the number of different ways joining neighboring letters to get the phrase DOS BANDOS is 1024.


You can record your path down as a sequence of 1s and 0s - left turn 0, right turn 1. Your record from any of the 4 D's at the top could be any combination of 8 ones/zeros. It gives you a total of 256 paths. Multiply it by 4 possible Ds to start with and you get the answer. Below is an example of a path record:




                D
/ left 0
O O
right 1
S S S
right 1
B B B B
/ left 0
A A A A A
right 1
N N N N N N
right 1
D D D D D D D
/ left 0
O O O O O O O O
/ left 0
S S S S S S S S S





share|improve this answer










New contributor




ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    Can you include a drawing in your answer so I could understand what were you referring with the $1$ and $0$ ?. Because I'm not getting the idea very clear. I believe there are zig zag options but I don't know how to account for them. Can you help with this part please?. :)
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell, I added an example of a recorded path, as you asked. I noticed Hugh has already helped you with understanding the answer. I hope the picture will also help clarifying it.
    $endgroup$
    – ppgdev
    1 hour ago














2












2








2





$begingroup$


The answer is - the number of different ways joining neighboring letters to get the phrase DOS BANDOS is 1024.


You can record your path down as a sequence of 1s and 0s - left turn 0, right turn 1. Your record from any of the 4 D's at the top could be any combination of 8 ones/zeros. It gives you a total of 256 paths. Multiply it by 4 possible Ds to start with and you get the answer. Below is an example of a path record:




                D
/ left 0
O O
right 1
S S S
right 1
B B B B
/ left 0
A A A A A
right 1
N N N N N N
right 1
D D D D D D D
/ left 0
O O O O O O O O
/ left 0
S S S S S S S S S





share|improve this answer










New contributor




ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




The answer is - the number of different ways joining neighboring letters to get the phrase DOS BANDOS is 1024.


You can record your path down as a sequence of 1s and 0s - left turn 0, right turn 1. Your record from any of the 4 D's at the top could be any combination of 8 ones/zeros. It gives you a total of 256 paths. Multiply it by 4 possible Ds to start with and you get the answer. Below is an example of a path record:




                D
/ left 0
O O
right 1
S S S
right 1
B B B B
/ left 0
A A A A A
right 1
N N N N N N
right 1
D D D D D D D
/ left 0
O O O O O O O O
/ left 0
S S S S S S S S S






share|improve this answer










New contributor




ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this answer



share|improve this answer








edited 1 hour ago





















New contributor




ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 2 hours ago









ppgdevppgdev

514




514




New contributor




ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






ppgdev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Can you include a drawing in your answer so I could understand what were you referring with the $1$ and $0$ ?. Because I'm not getting the idea very clear. I believe there are zig zag options but I don't know how to account for them. Can you help with this part please?. :)
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell, I added an example of a recorded path, as you asked. I noticed Hugh has already helped you with understanding the answer. I hope the picture will also help clarifying it.
    $endgroup$
    – ppgdev
    1 hour ago


















  • $begingroup$
    Can you include a drawing in your answer so I could understand what were you referring with the $1$ and $0$ ?. Because I'm not getting the idea very clear. I believe there are zig zag options but I don't know how to account for them. Can you help with this part please?. :)
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell, I added an example of a recorded path, as you asked. I noticed Hugh has already helped you with understanding the answer. I hope the picture will also help clarifying it.
    $endgroup$
    – ppgdev
    1 hour ago
















$begingroup$
Can you include a drawing in your answer so I could understand what were you referring with the $1$ and $0$ ?. Because I'm not getting the idea very clear. I believe there are zig zag options but I don't know how to account for them. Can you help with this part please?. :)
$endgroup$
– Chris Steinbeck Bell
2 hours ago




$begingroup$
Can you include a drawing in your answer so I could understand what were you referring with the $1$ and $0$ ?. Because I'm not getting the idea very clear. I believe there are zig zag options but I don't know how to account for them. Can you help with this part please?. :)
$endgroup$
– Chris Steinbeck Bell
2 hours ago












$begingroup$
@ChrisSteinbeckBell, I added an example of a recorded path, as you asked. I noticed Hugh has already helped you with understanding the answer. I hope the picture will also help clarifying it.
$endgroup$
– ppgdev
1 hour ago




$begingroup$
@ChrisSteinbeckBell, I added an example of a recorded path, as you asked. I noticed Hugh has already helped you with understanding the answer. I hope the picture will also help clarifying it.
$endgroup$
– ppgdev
1 hour ago











1












$begingroup$

The solution to this problem is...




$1024$


From any given circle that is not in the bottom row, you can either go down and to the left or down and to the right. Because there are four starting points and eight rows in which we make a turn, the solution is $4 times 2^{8} = 1024$.







share|improve this answer











$endgroup$













  • $begingroup$
    Precisely this is what I wanted to avoid. Can you explain why is it a power of $2$?. In my question what I felt was needed for me to understand the rationale behind this question was to accompany some drawing to justify an answer. Because I'm still stuck on how to get the answer.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell think about it this way: say you have one coin; it can be either "heads" or "tails" — that's $2$ possibilities. Now, add a second coin; now you can have "heads-heads", "heads-tails", "tails-heads", or "tails-tails" — that's $4$ possibilities. Now, add a third coin; now, you can have "heads-heads-heads", "heads-heads-tails", "heads-tails-heads", .... etc. for a total of $8$ possible arrangements. Notice that each time we add a new coin, the number of possibilities doubles [cont'd]
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    that's a more logical way!. But why each time a coin is added the result is multiplied and not a sum?. I mean why the possibilities do not just sum?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell [cont'd] this is because the new coin introduces another combination. Let me do an example. We already know that $3$ coins = $8$ possibilities, so let's try and mathematically calculate the number of combinations for $4$ coins. For each of the $8$ combinations with $3$ coins, we can add a "heads" to the end, or a "tails" to the end. That means that the number of possibilities doubles.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    To reference your comment as to why we multiply and not add, let me put it this way. Say that we're trying to calculate the number of possibilities for $10$ coins. Now, take a look at this picture, or this picture
    $endgroup$
    – Hugh
    2 hours ago


















1












$begingroup$

The solution to this problem is...




$1024$


From any given circle that is not in the bottom row, you can either go down and to the left or down and to the right. Because there are four starting points and eight rows in which we make a turn, the solution is $4 times 2^{8} = 1024$.







share|improve this answer











$endgroup$













  • $begingroup$
    Precisely this is what I wanted to avoid. Can you explain why is it a power of $2$?. In my question what I felt was needed for me to understand the rationale behind this question was to accompany some drawing to justify an answer. Because I'm still stuck on how to get the answer.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell think about it this way: say you have one coin; it can be either "heads" or "tails" — that's $2$ possibilities. Now, add a second coin; now you can have "heads-heads", "heads-tails", "tails-heads", or "tails-tails" — that's $4$ possibilities. Now, add a third coin; now, you can have "heads-heads-heads", "heads-heads-tails", "heads-tails-heads", .... etc. for a total of $8$ possible arrangements. Notice that each time we add a new coin, the number of possibilities doubles [cont'd]
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    that's a more logical way!. But why each time a coin is added the result is multiplied and not a sum?. I mean why the possibilities do not just sum?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell [cont'd] this is because the new coin introduces another combination. Let me do an example. We already know that $3$ coins = $8$ possibilities, so let's try and mathematically calculate the number of combinations for $4$ coins. For each of the $8$ combinations with $3$ coins, we can add a "heads" to the end, or a "tails" to the end. That means that the number of possibilities doubles.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    To reference your comment as to why we multiply and not add, let me put it this way. Say that we're trying to calculate the number of possibilities for $10$ coins. Now, take a look at this picture, or this picture
    $endgroup$
    – Hugh
    2 hours ago
















1












1








1





$begingroup$

The solution to this problem is...




$1024$


From any given circle that is not in the bottom row, you can either go down and to the left or down and to the right. Because there are four starting points and eight rows in which we make a turn, the solution is $4 times 2^{8} = 1024$.







share|improve this answer











$endgroup$



The solution to this problem is...




$1024$


From any given circle that is not in the bottom row, you can either go down and to the left or down and to the right. Because there are four starting points and eight rows in which we make a turn, the solution is $4 times 2^{8} = 1024$.








share|improve this answer














share|improve this answer



share|improve this answer








edited 2 hours ago

























answered 2 hours ago









HughHugh

2,0591923




2,0591923












  • $begingroup$
    Precisely this is what I wanted to avoid. Can you explain why is it a power of $2$?. In my question what I felt was needed for me to understand the rationale behind this question was to accompany some drawing to justify an answer. Because I'm still stuck on how to get the answer.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell think about it this way: say you have one coin; it can be either "heads" or "tails" — that's $2$ possibilities. Now, add a second coin; now you can have "heads-heads", "heads-tails", "tails-heads", or "tails-tails" — that's $4$ possibilities. Now, add a third coin; now, you can have "heads-heads-heads", "heads-heads-tails", "heads-tails-heads", .... etc. for a total of $8$ possible arrangements. Notice that each time we add a new coin, the number of possibilities doubles [cont'd]
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    that's a more logical way!. But why each time a coin is added the result is multiplied and not a sum?. I mean why the possibilities do not just sum?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell [cont'd] this is because the new coin introduces another combination. Let me do an example. We already know that $3$ coins = $8$ possibilities, so let's try and mathematically calculate the number of combinations for $4$ coins. For each of the $8$ combinations with $3$ coins, we can add a "heads" to the end, or a "tails" to the end. That means that the number of possibilities doubles.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    To reference your comment as to why we multiply and not add, let me put it this way. Say that we're trying to calculate the number of possibilities for $10$ coins. Now, take a look at this picture, or this picture
    $endgroup$
    – Hugh
    2 hours ago




















  • $begingroup$
    Precisely this is what I wanted to avoid. Can you explain why is it a power of $2$?. In my question what I felt was needed for me to understand the rationale behind this question was to accompany some drawing to justify an answer. Because I'm still stuck on how to get the answer.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell think about it this way: say you have one coin; it can be either "heads" or "tails" — that's $2$ possibilities. Now, add a second coin; now you can have "heads-heads", "heads-tails", "tails-heads", or "tails-tails" — that's $4$ possibilities. Now, add a third coin; now, you can have "heads-heads-heads", "heads-heads-tails", "heads-tails-heads", .... etc. for a total of $8$ possible arrangements. Notice that each time we add a new coin, the number of possibilities doubles [cont'd]
    $endgroup$
    – Hugh
    2 hours ago












  • $begingroup$
    that's a more logical way!. But why each time a coin is added the result is multiplied and not a sum?. I mean why the possibilities do not just sum?.
    $endgroup$
    – Chris Steinbeck Bell
    2 hours ago










  • $begingroup$
    @ChrisSteinbeckBell [cont'd] this is because the new coin introduces another combination. Let me do an example. We already know that $3$ coins = $8$ possibilities, so let's try and mathematically calculate the number of combinations for $4$ coins. For each of the $8$ combinations with $3$ coins, we can add a "heads" to the end, or a "tails" to the end. That means that the number of possibilities doubles.
    $endgroup$
    – Hugh
    2 hours ago










  • $begingroup$
    To reference your comment as to why we multiply and not add, let me put it this way. Say that we're trying to calculate the number of possibilities for $10$ coins. Now, take a look at this picture, or this picture
    $endgroup$
    – Hugh
    2 hours ago


















$begingroup$
Precisely this is what I wanted to avoid. Can you explain why is it a power of $2$?. In my question what I felt was needed for me to understand the rationale behind this question was to accompany some drawing to justify an answer. Because I'm still stuck on how to get the answer.
$endgroup$
– Chris Steinbeck Bell
2 hours ago




$begingroup$
Precisely this is what I wanted to avoid. Can you explain why is it a power of $2$?. In my question what I felt was needed for me to understand the rationale behind this question was to accompany some drawing to justify an answer. Because I'm still stuck on how to get the answer.
$endgroup$
– Chris Steinbeck Bell
2 hours ago












$begingroup$
@ChrisSteinbeckBell think about it this way: say you have one coin; it can be either "heads" or "tails" — that's $2$ possibilities. Now, add a second coin; now you can have "heads-heads", "heads-tails", "tails-heads", or "tails-tails" — that's $4$ possibilities. Now, add a third coin; now, you can have "heads-heads-heads", "heads-heads-tails", "heads-tails-heads", .... etc. for a total of $8$ possible arrangements. Notice that each time we add a new coin, the number of possibilities doubles [cont'd]
$endgroup$
– Hugh
2 hours ago






$begingroup$
@ChrisSteinbeckBell think about it this way: say you have one coin; it can be either "heads" or "tails" — that's $2$ possibilities. Now, add a second coin; now you can have "heads-heads", "heads-tails", "tails-heads", or "tails-tails" — that's $4$ possibilities. Now, add a third coin; now, you can have "heads-heads-heads", "heads-heads-tails", "heads-tails-heads", .... etc. for a total of $8$ possible arrangements. Notice that each time we add a new coin, the number of possibilities doubles [cont'd]
$endgroup$
– Hugh
2 hours ago














$begingroup$
that's a more logical way!. But why each time a coin is added the result is multiplied and not a sum?. I mean why the possibilities do not just sum?.
$endgroup$
– Chris Steinbeck Bell
2 hours ago




$begingroup$
that's a more logical way!. But why each time a coin is added the result is multiplied and not a sum?. I mean why the possibilities do not just sum?.
$endgroup$
– Chris Steinbeck Bell
2 hours ago












$begingroup$
@ChrisSteinbeckBell [cont'd] this is because the new coin introduces another combination. Let me do an example. We already know that $3$ coins = $8$ possibilities, so let's try and mathematically calculate the number of combinations for $4$ coins. For each of the $8$ combinations with $3$ coins, we can add a "heads" to the end, or a "tails" to the end. That means that the number of possibilities doubles.
$endgroup$
– Hugh
2 hours ago




$begingroup$
@ChrisSteinbeckBell [cont'd] this is because the new coin introduces another combination. Let me do an example. We already know that $3$ coins = $8$ possibilities, so let's try and mathematically calculate the number of combinations for $4$ coins. For each of the $8$ combinations with $3$ coins, we can add a "heads" to the end, or a "tails" to the end. That means that the number of possibilities doubles.
$endgroup$
– Hugh
2 hours ago












$begingroup$
To reference your comment as to why we multiply and not add, let me put it this way. Say that we're trying to calculate the number of possibilities for $10$ coins. Now, take a look at this picture, or this picture
$endgroup$
– Hugh
2 hours ago






$begingroup$
To reference your comment as to why we multiply and not add, let me put it this way. Say that we're trying to calculate the number of possibilities for $10$ coins. Now, take a look at this picture, or this picture
$endgroup$
– Hugh
2 hours ago




















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