Mixing
How many hexagonal paths?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Here is a hexagonal tiling, borrowed from Wikipedia.
I start in any hexagon on the left hand side. I end at any hexagon on the right hand side. I can only travel to the right, not up, down or backwards.
In how many ways can this be done?
combinatorics
asked 1 hour ago
JonMark Perry
15.3k52973
add a comment |Â
up vote
4
down vote
favorite
Here is a hexagonal tiling, borrowed from Wikipedia.
I start in any hexagon on the left hand side. I end at any hexagon on the right hand side. I can only travel to the right, not up, down or backwards.
In how many ways can this be done?
combinatorics
asked 1 hour ago
JonMark Perry
15.3k52973
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Here is a hexagonal tiling, borrowed from Wikipedia.
I start in any hexagon on the left hand side. I end at any hexagon on the right hand side. I can only travel to the right, not up, down or backwards.
In how many ways can this be done?
combinatorics
asked 1 hour ago
JonMark Perry
15.3k52973
Here is a hexagonal tiling, borrowed from Wikipedia.
I start in any hexagon on the left hand side. I end at any hexagon on the right hand side. I can only travel to the right, not up, down or backwards.
In how many ways can this be done?
combinatorics
combinatorics
asked 1 hour ago
JonMark Perry
15.3k52973
asked 1 hour ago
JonMark Perry
15.3k52973
asked 1 hour ago
JonMark Perry
15.3k52973
asked 1 hour ago
JonMark Perry
15.3k52973
asked 1 hour ago
JonMark Perry
15.3k52973
15.3k52973
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
The correct answer might be:
73392?
Reasoning:
Well, I made a Pascal's Triangle-like chart with alternating 11 and 12 columns, and 14 rows, where each number was the sum of the two numbers northwest and northeast of it. I added all of the numbers on the bottom row. It's quite likely I messed up some addition, though.
answered 53 mins ago
Excited Raichu
2,506231
The method seems to be correct, but I'm afraid your calculation is flawed indeed.
– elias
17 mins ago
@elias yeah, I thought that would be the case.
– Excited Raichu
16 mins ago
Also I'm sure you meant northwest and southwest instead of northeast.
– elias
15 mins ago
1
@elias I did the chart vertically instead of horizontally like the question asked (same result, and easier for me to draw), so for me it was northeast. I see how this could cause a bit of confusion though.
– Excited Raichu
11 mins ago
add a comment |Â
up vote
3
down vote
The answer is
74280
Method
Every cell in the first column can be reached 1 way.
Other cells can be reached as many different ways, as the sum of their two left neighbours, giving:
The sum of the numbers in the last column is the answer to the question posted.
answered 20 mins ago
elias
8,25432151
add a comment |Â
up vote
0
down vote
84652?
There are 11 hexagons on the left hand side and 13 steps until you reach the end.
1st step: Two possibilities for each tile, so 11·2=22 paths.
2nd step: 2 of the previous paths reach the top or the bottom of the tiling so there's only one possible move there, the rest of them can move to two hexagons: (22 - 2)·2 + 2 = 42 paths.
3rd step: You can move to two different hexagons on each path so it*s 42·2 = 84 paths.
...
Following this method you end up with 42326 possible paths at the second to last column, and since there are two possible choices for each path, 42326·2 = 84652 paths.
answered 1 hour ago
NudgeNudge
529113
1
I'm currently also writing an answer, and that would be correct, if the set up constantly grew in height as it moved right. However, the top and bottom hexes of the even columns only have 1 destination, so it is actually less than this.
– AHKieran
58 mins ago
You're right, thanks for the correction! I've recalculated, though I'm probably still missing something.
– NudgeNudge
34 mins ago
1
You and I have the same start, but vastly different endings xD
– AHKieran
28 mins ago
add a comment |Â
up vote
0
down vote
Answer:
$6,564,516$
Method:
Consider the first 2 columns of this diagram. There are 11 starting positions, and 12 destinations. Each starting position can go to one of 2 destinations, so the number of possible paths is equal to $11 cdot 2 = 22$.
Now, consider columns 2 and 3 of the diagram. There are 12 starting positions, and 11 destinations. 10 of the starting positions have 2 possible destinations, and 2 of them only have 1 possible destination (the top and bottom ones). This means the total number of possible paths is $(10 cdot 2) + 2 = 22$.
If we now consider the first 3 columns, it can be seen that each of the internal starting positions has 4 possible paths, and the top and bottom ones have 3. Therefore, in total, there are $(9 cdot 4) + (2 cdot 3) = 36 + 6 = 42$ possible paths across the first 3 columns.
In the initial consideration, the inner hexes were each finished on twice, and the outer ones once. Therefore, a second way to work out the number of paths in the third consideration, would be to edit the calculation for consideration 2 to be $((10 cdot 2) cdot 2) + (2 cdot 1) = 40 + 2 = 42$.
Next, considering the first 4 columns, each of the paths already determined for the first 3, has 2 more possible destinations, so the number of paths would total $42 cdot 2 = 84$.
Using these values, I can determine a formula for calculating the number of paths, where n increases by 1 for every repeating pair of columns.
$Sigma_n = ((Sigma_n-1 - n) cdot 2n) + 2n$
which can be simplified to
$Sigma_n = (Sigma_n-1 + 1 - n) cdot 2n$
Therefore, we can step through this calculation to determine the results of:
$Sigma_0 = 11$
$Sigma_1 = 22$
$Sigma_2 = 84$
$Sigma_3 = 492$
$Sigma_4 = 3912$
$Sigma_5 = 39080$
$Sigma_6 = 468,900$
$Sigma_7 = 6,564,516$
As there are 7 pairs of columns in the above image, (if my formula is correct), I believe there to be $6,564,516$ possible paths.
answered 29 mins ago
AHKieran
3,084624
2
I'm pretty sure @NudgeNudge 's earlier answer of 90112 is an upper bound for the answer.
– Excited Raichu
19 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The correct answer might be:
73392?
Reasoning:
Well, I made a Pascal's Triangle-like chart with alternating 11 and 12 columns, and 14 rows, where each number was the sum of the two numbers northwest and northeast of it. I added all of the numbers on the bottom row. It's quite likely I messed up some addition, though.
answered 53 mins ago
Excited Raichu
2,506231
The method seems to be correct, but I'm afraid your calculation is flawed indeed.
– elias
17 mins ago
@elias yeah, I thought that would be the case.
– Excited Raichu
16 mins ago
Also I'm sure you meant northwest and southwest instead of northeast.
– elias
15 mins ago
1
@elias I did the chart vertically instead of horizontally like the question asked (same result, and easier for me to draw), so for me it was northeast. I see how this could cause a bit of confusion though.
– Excited Raichu
11 mins ago
add a comment |Â
up vote
3
down vote
The correct answer might be:
73392?
Reasoning:
Well, I made a Pascal's Triangle-like chart with alternating 11 and 12 columns, and 14 rows, where each number was the sum of the two numbers northwest and northeast of it. I added all of the numbers on the bottom row. It's quite likely I messed up some addition, though.
answered 53 mins ago
Excited Raichu
2,506231
The method seems to be correct, but I'm afraid your calculation is flawed indeed.
– elias
17 mins ago
@elias yeah, I thought that would be the case.
– Excited Raichu
16 mins ago
Also I'm sure you meant northwest and southwest instead of northeast.
– elias
15 mins ago
1
@elias I did the chart vertically instead of horizontally like the question asked (same result, and easier for me to draw), so for me it was northeast. I see how this could cause a bit of confusion though.
– Excited Raichu
11 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The correct answer might be:
73392?
Reasoning:
Well, I made a Pascal's Triangle-like chart with alternating 11 and 12 columns, and 14 rows, where each number was the sum of the two numbers northwest and northeast of it. I added all of the numbers on the bottom row. It's quite likely I messed up some addition, though.
answered 53 mins ago
Excited Raichu
2,506231
The correct answer might be:
73392?
Reasoning:
Well, I made a Pascal's Triangle-like chart with alternating 11 and 12 columns, and 14 rows, where each number was the sum of the two numbers northwest and northeast of it. I added all of the numbers on the bottom row. It's quite likely I messed up some addition, though.
answered 53 mins ago
Excited Raichu
2,506231
answered 53 mins ago
Excited Raichu
2,506231
answered 53 mins ago
Excited Raichu
2,506231
answered 53 mins ago
Excited Raichu
2,506231
2,506231
The method seems to be correct, but I'm afraid your calculation is flawed indeed.
– elias
17 mins ago
@elias yeah, I thought that would be the case.
– Excited Raichu
16 mins ago
Also I'm sure you meant northwest and southwest instead of northeast.
– elias
15 mins ago
1
@elias I did the chart vertically instead of horizontally like the question asked (same result, and easier for me to draw), so for me it was northeast. I see how this could cause a bit of confusion though.
– Excited Raichu
11 mins ago
add a comment |Â
The method seems to be correct, but I'm afraid your calculation is flawed indeed.
– elias
17 mins ago
@elias yeah, I thought that would be the case.
– Excited Raichu
16 mins ago
Also I'm sure you meant northwest and southwest instead of northeast.
– elias
15 mins ago
1
@elias I did the chart vertically instead of horizontally like the question asked (same result, and easier for me to draw), so for me it was northeast. I see how this could cause a bit of confusion though.
– Excited Raichu
11 mins ago
The method seems to be correct, but I'm afraid your calculation is flawed indeed.
– elias
17 mins ago
The method seems to be correct, but I'm afraid your calculation is flawed indeed.
– elias
17 mins ago
@elias yeah, I thought that would be the case.
– Excited Raichu
16 mins ago
@elias yeah, I thought that would be the case.
– Excited Raichu
16 mins ago
Also I'm sure you meant northwest and southwest instead of northeast.
– elias
15 mins ago
Also I'm sure you meant northwest and southwest instead of northeast.
– elias
15 mins ago
1
1
@elias I did the chart vertically instead of horizontally like the question asked (same result, and easier for me to draw), so for me it was northeast. I see how this could cause a bit of confusion though.
– Excited Raichu
11 mins ago
@elias I did the chart vertically instead of horizontally like the question asked (same result, and easier for me to draw), so for me it was northeast. I see how this could cause a bit of confusion though.
– Excited Raichu
11 mins ago
add a comment |Â
up vote
3
down vote
The answer is
74280
Method
Every cell in the first column can be reached 1 way.
Other cells can be reached as many different ways, as the sum of their two left neighbours, giving:
The sum of the numbers in the last column is the answer to the question posted.
answered 20 mins ago
elias
8,25432151
add a comment |Â
up vote
3
down vote
The answer is
74280
Method
Every cell in the first column can be reached 1 way.
Other cells can be reached as many different ways, as the sum of their two left neighbours, giving:
The sum of the numbers in the last column is the answer to the question posted.
answered 20 mins ago
elias
8,25432151
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The answer is
74280
Method
Every cell in the first column can be reached 1 way.
Other cells can be reached as many different ways, as the sum of their two left neighbours, giving:
The sum of the numbers in the last column is the answer to the question posted.
answered 20 mins ago
elias
8,25432151
The answer is
74280
Method
Every cell in the first column can be reached 1 way.
Other cells can be reached as many different ways, as the sum of their two left neighbours, giving:
The sum of the numbers in the last column is the answer to the question posted.
answered 20 mins ago
elias
8,25432151
answered 20 mins ago
elias
8,25432151
answered 20 mins ago
elias
8,25432151
answered 20 mins ago
elias
8,25432151
8,25432151
add a comment |Â
add a comment |Â
up vote
0
down vote
84652?
There are 11 hexagons on the left hand side and 13 steps until you reach the end.
1st step: Two possibilities for each tile, so 11·2=22 paths.
2nd step: 2 of the previous paths reach the top or the bottom of the tiling so there's only one possible move there, the rest of them can move to two hexagons: (22 - 2)·2 + 2 = 42 paths.
3rd step: You can move to two different hexagons on each path so it*s 42·2 = 84 paths.
...
Following this method you end up with 42326 possible paths at the second to last column, and since there are two possible choices for each path, 42326·2 = 84652 paths.
answered 1 hour ago
NudgeNudge
529113
1
I'm currently also writing an answer, and that would be correct, if the set up constantly grew in height as it moved right. However, the top and bottom hexes of the even columns only have 1 destination, so it is actually less than this.
– AHKieran
58 mins ago
You're right, thanks for the correction! I've recalculated, though I'm probably still missing something.
– NudgeNudge
34 mins ago
1
You and I have the same start, but vastly different endings xD
– AHKieran
28 mins ago
add a comment |Â
up vote
0
down vote
84652?
There are 11 hexagons on the left hand side and 13 steps until you reach the end.
1st step: Two possibilities for each tile, so 11·2=22 paths.
2nd step: 2 of the previous paths reach the top or the bottom of the tiling so there's only one possible move there, the rest of them can move to two hexagons: (22 - 2)·2 + 2 = 42 paths.
3rd step: You can move to two different hexagons on each path so it*s 42·2 = 84 paths.
...
Following this method you end up with 42326 possible paths at the second to last column, and since there are two possible choices for each path, 42326·2 = 84652 paths.
answered 1 hour ago
NudgeNudge
529113
1
I'm currently also writing an answer, and that would be correct, if the set up constantly grew in height as it moved right. However, the top and bottom hexes of the even columns only have 1 destination, so it is actually less than this.
– AHKieran
58 mins ago
You're right, thanks for the correction! I've recalculated, though I'm probably still missing something.
– NudgeNudge
34 mins ago
1
You and I have the same start, but vastly different endings xD
– AHKieran
28 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
84652?
There are 11 hexagons on the left hand side and 13 steps until you reach the end.
1st step: Two possibilities for each tile, so 11·2=22 paths.
2nd step: 2 of the previous paths reach the top or the bottom of the tiling so there's only one possible move there, the rest of them can move to two hexagons: (22 - 2)·2 + 2 = 42 paths.
3rd step: You can move to two different hexagons on each path so it*s 42·2 = 84 paths.
...
Following this method you end up with 42326 possible paths at the second to last column, and since there are two possible choices for each path, 42326·2 = 84652 paths.
answered 1 hour ago
NudgeNudge
529113
84652?
There are 11 hexagons on the left hand side and 13 steps until you reach the end.
1st step: Two possibilities for each tile, so 11·2=22 paths.
2nd step: 2 of the previous paths reach the top or the bottom of the tiling so there's only one possible move there, the rest of them can move to two hexagons: (22 - 2)·2 + 2 = 42 paths.
3rd step: You can move to two different hexagons on each path so it*s 42·2 = 84 paths.
...
Following this method you end up with 42326 possible paths at the second to last column, and since there are two possible choices for each path, 42326·2 = 84652 paths.
answered 1 hour ago
NudgeNudge
529113
edited 34 mins ago
answered 1 hour ago
NudgeNudge
529113
answered 1 hour ago
NudgeNudge
529113
answered 1 hour ago
NudgeNudge
529113
529113
1
I'm currently also writing an answer, and that would be correct, if the set up constantly grew in height as it moved right. However, the top and bottom hexes of the even columns only have 1 destination, so it is actually less than this.
– AHKieran
58 mins ago
You're right, thanks for the correction! I've recalculated, though I'm probably still missing something.
– NudgeNudge
34 mins ago
1
You and I have the same start, but vastly different endings xD
– AHKieran
28 mins ago
add a comment |Â
1
I'm currently also writing an answer, and that would be correct, if the set up constantly grew in height as it moved right. However, the top and bottom hexes of the even columns only have 1 destination, so it is actually less than this.
– AHKieran
58 mins ago
You're right, thanks for the correction! I've recalculated, though I'm probably still missing something.
– NudgeNudge
34 mins ago
1
You and I have the same start, but vastly different endings xD
– AHKieran
28 mins ago
1
1
I'm currently also writing an answer, and that would be correct, if the set up constantly grew in height as it moved right. However, the top and bottom hexes of the even columns only have 1 destination, so it is actually less than this.
– AHKieran
58 mins ago
I'm currently also writing an answer, and that would be correct, if the set up constantly grew in height as it moved right. However, the top and bottom hexes of the even columns only have 1 destination, so it is actually less than this.
– AHKieran
58 mins ago
You're right, thanks for the correction! I've recalculated, though I'm probably still missing something.
– NudgeNudge
34 mins ago
You're right, thanks for the correction! I've recalculated, though I'm probably still missing something.
– NudgeNudge
34 mins ago
1
1
You and I have the same start, but vastly different endings xD
– AHKieran
28 mins ago
You and I have the same start, but vastly different endings xD
– AHKieran
28 mins ago
add a comment |Â
up vote
0
down vote
Answer:
$6,564,516$
Method:
Consider the first 2 columns of this diagram. There are 11 starting positions, and 12 destinations. Each starting position can go to one of 2 destinations, so the number of possible paths is equal to $11 cdot 2 = 22$.
Now, consider columns 2 and 3 of the diagram. There are 12 starting positions, and 11 destinations. 10 of the starting positions have 2 possible destinations, and 2 of them only have 1 possible destination (the top and bottom ones). This means the total number of possible paths is $(10 cdot 2) + 2 = 22$.
If we now consider the first 3 columns, it can be seen that each of the internal starting positions has 4 possible paths, and the top and bottom ones have 3. Therefore, in total, there are $(9 cdot 4) + (2 cdot 3) = 36 + 6 = 42$ possible paths across the first 3 columns.
In the initial consideration, the inner hexes were each finished on twice, and the outer ones once. Therefore, a second way to work out the number of paths in the third consideration, would be to edit the calculation for consideration 2 to be $((10 cdot 2) cdot 2) + (2 cdot 1) = 40 + 2 = 42$.
Next, considering the first 4 columns, each of the paths already determined for the first 3, has 2 more possible destinations, so the number of paths would total $42 cdot 2 = 84$.
Using these values, I can determine a formula for calculating the number of paths, where n increases by 1 for every repeating pair of columns.
$Sigma_n = ((Sigma_n-1 - n) cdot 2n) + 2n$
which can be simplified to
$Sigma_n = (Sigma_n-1 + 1 - n) cdot 2n$
Therefore, we can step through this calculation to determine the results of:
$Sigma_0 = 11$
$Sigma_1 = 22$
$Sigma_2 = 84$
$Sigma_3 = 492$
$Sigma_4 = 3912$
$Sigma_5 = 39080$
$Sigma_6 = 468,900$
$Sigma_7 = 6,564,516$
As there are 7 pairs of columns in the above image, (if my formula is correct), I believe there to be $6,564,516$ possible paths.
answered 29 mins ago
AHKieran
3,084624
2
I'm pretty sure @NudgeNudge 's earlier answer of 90112 is an upper bound for the answer.
– Excited Raichu
19 mins ago
add a comment |Â
up vote
0
down vote
Answer:
$6,564,516$
Method:
Consider the first 2 columns of this diagram. There are 11 starting positions, and 12 destinations. Each starting position can go to one of 2 destinations, so the number of possible paths is equal to $11 cdot 2 = 22$.
Now, consider columns 2 and 3 of the diagram. There are 12 starting positions, and 11 destinations. 10 of the starting positions have 2 possible destinations, and 2 of them only have 1 possible destination (the top and bottom ones). This means the total number of possible paths is $(10 cdot 2) + 2 = 22$.
If we now consider the first 3 columns, it can be seen that each of the internal starting positions has 4 possible paths, and the top and bottom ones have 3. Therefore, in total, there are $(9 cdot 4) + (2 cdot 3) = 36 + 6 = 42$ possible paths across the first 3 columns.
In the initial consideration, the inner hexes were each finished on twice, and the outer ones once. Therefore, a second way to work out the number of paths in the third consideration, would be to edit the calculation for consideration 2 to be $((10 cdot 2) cdot 2) + (2 cdot 1) = 40 + 2 = 42$.
Next, considering the first 4 columns, each of the paths already determined for the first 3, has 2 more possible destinations, so the number of paths would total $42 cdot 2 = 84$.
Using these values, I can determine a formula for calculating the number of paths, where n increases by 1 for every repeating pair of columns.
$Sigma_n = ((Sigma_n-1 - n) cdot 2n) + 2n$
which can be simplified to
$Sigma_n = (Sigma_n-1 + 1 - n) cdot 2n$
Therefore, we can step through this calculation to determine the results of:
$Sigma_0 = 11$
$Sigma_1 = 22$
$Sigma_2 = 84$
$Sigma_3 = 492$
$Sigma_4 = 3912$
$Sigma_5 = 39080$
$Sigma_6 = 468,900$
$Sigma_7 = 6,564,516$
As there are 7 pairs of columns in the above image, (if my formula is correct), I believe there to be $6,564,516$ possible paths.
answered 29 mins ago
AHKieran
3,084624
2
I'm pretty sure @NudgeNudge 's earlier answer of 90112 is an upper bound for the answer.
– Excited Raichu
19 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Answer:
$6,564,516$
Method:
Consider the first 2 columns of this diagram. There are 11 starting positions, and 12 destinations. Each starting position can go to one of 2 destinations, so the number of possible paths is equal to $11 cdot 2 = 22$.
Now, consider columns 2 and 3 of the diagram. There are 12 starting positions, and 11 destinations. 10 of the starting positions have 2 possible destinations, and 2 of them only have 1 possible destination (the top and bottom ones). This means the total number of possible paths is $(10 cdot 2) + 2 = 22$.
If we now consider the first 3 columns, it can be seen that each of the internal starting positions has 4 possible paths, and the top and bottom ones have 3. Therefore, in total, there are $(9 cdot 4) + (2 cdot 3) = 36 + 6 = 42$ possible paths across the first 3 columns.
In the initial consideration, the inner hexes were each finished on twice, and the outer ones once. Therefore, a second way to work out the number of paths in the third consideration, would be to edit the calculation for consideration 2 to be $((10 cdot 2) cdot 2) + (2 cdot 1) = 40 + 2 = 42$.
Next, considering the first 4 columns, each of the paths already determined for the first 3, has 2 more possible destinations, so the number of paths would total $42 cdot 2 = 84$.
Using these values, I can determine a formula for calculating the number of paths, where n increases by 1 for every repeating pair of columns.
$Sigma_n = ((Sigma_n-1 - n) cdot 2n) + 2n$
which can be simplified to
$Sigma_n = (Sigma_n-1 + 1 - n) cdot 2n$
Therefore, we can step through this calculation to determine the results of:
$Sigma_0 = 11$
$Sigma_1 = 22$
$Sigma_2 = 84$
$Sigma_3 = 492$
$Sigma_4 = 3912$
$Sigma_5 = 39080$
$Sigma_6 = 468,900$
$Sigma_7 = 6,564,516$
As there are 7 pairs of columns in the above image, (if my formula is correct), I believe there to be $6,564,516$ possible paths.
answered 29 mins ago
AHKieran
3,084624
Answer:
$6,564,516$
Method:
Consider the first 2 columns of this diagram. There are 11 starting positions, and 12 destinations. Each starting position can go to one of 2 destinations, so the number of possible paths is equal to $11 cdot 2 = 22$.
Now, consider columns 2 and 3 of the diagram. There are 12 starting positions, and 11 destinations. 10 of the starting positions have 2 possible destinations, and 2 of them only have 1 possible destination (the top and bottom ones). This means the total number of possible paths is $(10 cdot 2) + 2 = 22$.
If we now consider the first 3 columns, it can be seen that each of the internal starting positions has 4 possible paths, and the top and bottom ones have 3. Therefore, in total, there are $(9 cdot 4) + (2 cdot 3) = 36 + 6 = 42$ possible paths across the first 3 columns.
In the initial consideration, the inner hexes were each finished on twice, and the outer ones once. Therefore, a second way to work out the number of paths in the third consideration, would be to edit the calculation for consideration 2 to be $((10 cdot 2) cdot 2) + (2 cdot 1) = 40 + 2 = 42$.
Next, considering the first 4 columns, each of the paths already determined for the first 3, has 2 more possible destinations, so the number of paths would total $42 cdot 2 = 84$.
Using these values, I can determine a formula for calculating the number of paths, where n increases by 1 for every repeating pair of columns.
$Sigma_n = ((Sigma_n-1 - n) cdot 2n) + 2n$
which can be simplified to
$Sigma_n = (Sigma_n-1 + 1 - n) cdot 2n$
Therefore, we can step through this calculation to determine the results of:
$Sigma_0 = 11$
$Sigma_1 = 22$
$Sigma_2 = 84$
$Sigma_3 = 492$
$Sigma_4 = 3912$
$Sigma_5 = 39080$
$Sigma_6 = 468,900$
$Sigma_7 = 6,564,516$
As there are 7 pairs of columns in the above image, (if my formula is correct), I believe there to be $6,564,516$ possible paths.
answered 29 mins ago
AHKieran
3,084624
answered 29 mins ago
AHKieran
3,084624
answered 29 mins ago
AHKieran
3,084624
answered 29 mins ago
AHKieran
3,084624
3,084624
2
I'm pretty sure @NudgeNudge 's earlier answer of 90112 is an upper bound for the answer.
– Excited Raichu
19 mins ago
add a comment |Â
2
I'm pretty sure @NudgeNudge 's earlier answer of 90112 is an upper bound for the answer.
– Excited Raichu
19 mins ago
2
2
I'm pretty sure @NudgeNudge 's earlier answer of 90112 is an upper bound for the answer.
– Excited Raichu
19 mins ago
I'm pretty sure @NudgeNudge 's earlier answer of 90112 is an upper bound for the answer.
– Excited Raichu
19 mins ago
add a comment |Â
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