Move fast … Or you will loseWho can find the most efficient path counting algorithm?The Temple Of Demdarr,...
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Move fast … Or you will lose
Who can find the most efficient path counting algorithm?The Temple Of Demdarr, God Of Puzzling (Part 1)How many paths are there?How many ways to pick the fruitHow many ways to pick the fruit 2Counting ant movementsWow, Puzzle. Woah, NumbersKnight's Tour QuestionThe King's Routes Problem: How many possibilities?Find a specific path on an n x n grid
$begingroup$
Suppose you're on a 4 × 6 grid, and want to go from the bottom left to the top right. How many different paths can you take? Avoid backtracking -- you can only move right or up.
pattern combinatorics
edited 1 hour ago
JonMark Perry
19.3k63991
asked 12 hours ago
Alpha-qAlpha-q
343
$endgroup$
add a comment |
$begingroup$
Suppose you're on a 4 × 6 grid, and want to go from the bottom left to the top right. How many different paths can you take? Avoid backtracking -- you can only move right or up.
pattern combinatorics
edited 1 hour ago
JonMark Perry
19.3k63991
asked 12 hours ago
Alpha-qAlpha-q
343
$endgroup$
2
$begingroup$
I think its from this site betterexplained.com/articles/…
$endgroup$
– Purple
11 hours ago
2
$begingroup$
(In the future please be aware that for content you did not create yourself, proper attribution is required. You need to include (at minimum) where it came from—and any additional context you can provide is often helpful to solvers. Posts which use someone else's content without attribution are generally deleted.)
$endgroup$
– Rubio♦
9 hours ago
add a comment |
$begingroup$
Suppose you're on a 4 × 6 grid, and want to go from the bottom left to the top right. How many different paths can you take? Avoid backtracking -- you can only move right or up.
pattern combinatorics
edited 1 hour ago
JonMark Perry
19.3k63991
asked 12 hours ago
Alpha-qAlpha-q
343
$endgroup$
Suppose you're on a 4 × 6 grid, and want to go from the bottom left to the top right. How many different paths can you take? Avoid backtracking -- you can only move right or up.
pattern combinatorics
pattern combinatorics
edited 1 hour ago
JonMark Perry
19.3k63991
asked 12 hours ago
Alpha-qAlpha-q
343
edited 1 hour ago
JonMark Perry
19.3k63991
asked 12 hours ago
Alpha-qAlpha-q
343
edited 1 hour ago
JonMark Perry
19.3k63991
edited 1 hour ago
JonMark Perry
19.3k63991
edited 1 hour ago
JonMark Perry
19.3k63991
19.3k63991
asked 12 hours ago
Alpha-qAlpha-q
343
asked 12 hours ago
Alpha-qAlpha-q
343
asked 12 hours ago
Alpha-qAlpha-q
343
343
2
$begingroup$
I think its from this site betterexplained.com/articles/…
$endgroup$
– Purple
11 hours ago
2
$begingroup$
(In the future please be aware that for content you did not create yourself, proper attribution is required. You need to include (at minimum) where it came from—and any additional context you can provide is often helpful to solvers. Posts which use someone else's content without attribution are generally deleted.)
$endgroup$
– Rubio♦
9 hours ago
add a comment |
2
$begingroup$
I think its from this site betterexplained.com/articles/…
$endgroup$
– Purple
11 hours ago
2
$begingroup$
(In the future please be aware that for content you did not create yourself, proper attribution is required. You need to include (at minimum) where it came from—and any additional context you can provide is often helpful to solvers. Posts which use someone else's content without attribution are generally deleted.)
$endgroup$
– Rubio♦
9 hours ago
2
2
$begingroup$
I think its from this site betterexplained.com/articles/…
$endgroup$
– Purple
11 hours ago
$begingroup$
I think its from this site betterexplained.com/articles/…
$endgroup$
– Purple
11 hours ago
2
2
$begingroup$
(In the future please be aware that for content you did not create yourself, proper attribution is required. You need to include (at minimum) where it came from—and any additional context you can provide is often helpful to solvers. Posts which use someone else's content without attribution are generally deleted.)
$endgroup$
– Rubio♦
9 hours ago
$begingroup$
(In the future please be aware that for content you did not create yourself, proper attribution is required. You need to include (at minimum) where it came from—and any additional context you can provide is often helpful to solvers. Posts which use someone else's content without attribution are generally deleted.)
$endgroup$
– Rubio♦
9 hours ago
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Instead of having 6 rights at 4 ups, imagine we start with 10 rights (r r r r r r r r r r).
Clearly this won't do: we need to change 4 of those rights into ups. How many ways can we pick 4 rights to change?
Well, we have 10 choices for the first 'right' to convert (see the combinations article). And 9 for the second, 8 for the third, and 7 choices for the final right-to-up conversion. There are 10 * 9 * 8 * 7 = 10!/6! = 5040 possibilities.
But, wait! We need to remove the redundancies: after all, converting moves #1 #2 #3 and #4 (in that order) is the same as converting #4 #3 #2 #1. We have 4! (4 * 3 * 2 * 1 = 24) ways to rearrange the ups we picked, so we finally get:
displaystyle{frac{(10!/6!)}{4!} = frac{5040}{24} = 210 }
We're just picking the items to convert (10!/6!) and dividing out the redundancies (4!).
Source of answere :https://betterexplained.com/articles/navigate-a-grid-using-combinations-and-permutations/
answered 38 mins ago
PurplePurple
726113
$endgroup$
1
$begingroup$
You should cite your source. This is clearly copied directly from the page you linked to in your comment above. Even the image is copied.
$endgroup$
– Daniel Mathias
14 mins ago
$begingroup$
@DanielMathias sorry sir I had made a mistake. I didn’t know that I should should provide link so a humble sorry . But the truth to be told mine answere was the same but lacked some explanation and a image , plz take back your devote . To support me as I am a beginner.
$endgroup$
– Purple
1 min ago
add a comment |
$begingroup$
This is, I'm sure, answered somewhere else. It is also related to Pascal's triangle.
Simply fill out the grid as follows:
In this grid, each number represents the number of ways of getting to that particular intersection. And that number is precisely the number of ways to get to the intersection below it added to the number of ways to get to the intersection to the left of it.
answered 11 hours ago
Dr XorileDr Xorile
12.8k22569
$endgroup$
add a comment |
$begingroup$
A more mathematically oriented answer:
You have $10$ moves to make in total and you need to choose which $4$ of them are going to be up.
The number of ways to do that is $${10choose 4}=210$$
answered 4 hours ago
Arnaud MortierArnaud Mortier
688211
$endgroup$
add a comment |
$begingroup$
Seems complicated but , but I think this may help even though there are some good answer
There are ten ways to sum up .choose any four of them which are directing upwards
answered 1 hour ago
HavookHavook
212
New contributor
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Write them out. Start with
rrrrrruuuu
rrrrruruuu
rrrrruuruu
...
uuuurrrrrr
and then count them up. You should have 210 in total.
Or:
$$binom30+binom41+binom52+binom63+binom74+binom85+binom96$$
$$=1+4+10+20+35+56+84$$
$$=210$$
answered 40 mins ago
JonMark PerryJonMark Perry
19.3k63991
$endgroup$
add a comment |
$begingroup$
YIPPY KAPPA IDDY GANGWA .........................................................................................................................☻☻☻☻☻☻☻☻☻☻☻☻☻☻
answered 1 hour ago
GangwaGangwa
1
New contributor
Gangwa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Instead of having 6 rights at 4 ups, imagine we start with 10 rights (r r r r r r r r r r).
Clearly this won't do: we need to change 4 of those rights into ups. How many ways can we pick 4 rights to change?
Well, we have 10 choices for the first 'right' to convert (see the combinations article). And 9 for the second, 8 for the third, and 7 choices for the final right-to-up conversion. There are 10 * 9 * 8 * 7 = 10!/6! = 5040 possibilities.
But, wait! We need to remove the redundancies: after all, converting moves #1 #2 #3 and #4 (in that order) is the same as converting #4 #3 #2 #1. We have 4! (4 * 3 * 2 * 1 = 24) ways to rearrange the ups we picked, so we finally get:
displaystyle{frac{(10!/6!)}{4!} = frac{5040}{24} = 210 }
We're just picking the items to convert (10!/6!) and dividing out the redundancies (4!).
Source of answere :https://betterexplained.com/articles/navigate-a-grid-using-combinations-and-permutations/
answered 38 mins ago
PurplePurple
726113
$endgroup$
1
$begingroup$
You should cite your source. This is clearly copied directly from the page you linked to in your comment above. Even the image is copied.
$endgroup$
– Daniel Mathias
14 mins ago
$begingroup$
@DanielMathias sorry sir I had made a mistake. I didn’t know that I should should provide link so a humble sorry . But the truth to be told mine answere was the same but lacked some explanation and a image , plz take back your devote . To support me as I am a beginner.
$endgroup$
– Purple
1 min ago
add a comment |
$begingroup$
Instead of having 6 rights at 4 ups, imagine we start with 10 rights (r r r r r r r r r r).
Clearly this won't do: we need to change 4 of those rights into ups. How many ways can we pick 4 rights to change?
Well, we have 10 choices for the first 'right' to convert (see the combinations article). And 9 for the second, 8 for the third, and 7 choices for the final right-to-up conversion. There are 10 * 9 * 8 * 7 = 10!/6! = 5040 possibilities.
But, wait! We need to remove the redundancies: after all, converting moves #1 #2 #3 and #4 (in that order) is the same as converting #4 #3 #2 #1. We have 4! (4 * 3 * 2 * 1 = 24) ways to rearrange the ups we picked, so we finally get:
displaystyle{frac{(10!/6!)}{4!} = frac{5040}{24} = 210 }
We're just picking the items to convert (10!/6!) and dividing out the redundancies (4!).
Source of answere :https://betterexplained.com/articles/navigate-a-grid-using-combinations-and-permutations/
answered 38 mins ago
PurplePurple
726113
$endgroup$
1
$begingroup$
You should cite your source. This is clearly copied directly from the page you linked to in your comment above. Even the image is copied.
$endgroup$
– Daniel Mathias
14 mins ago
$begingroup$
@DanielMathias sorry sir I had made a mistake. I didn’t know that I should should provide link so a humble sorry . But the truth to be told mine answere was the same but lacked some explanation and a image , plz take back your devote . To support me as I am a beginner.
$endgroup$
– Purple
1 min ago
add a comment |
$begingroup$
Instead of having 6 rights at 4 ups, imagine we start with 10 rights (r r r r r r r r r r).
Clearly this won't do: we need to change 4 of those rights into ups. How many ways can we pick 4 rights to change?
Well, we have 10 choices for the first 'right' to convert (see the combinations article). And 9 for the second, 8 for the third, and 7 choices for the final right-to-up conversion. There are 10 * 9 * 8 * 7 = 10!/6! = 5040 possibilities.
But, wait! We need to remove the redundancies: after all, converting moves #1 #2 #3 and #4 (in that order) is the same as converting #4 #3 #2 #1. We have 4! (4 * 3 * 2 * 1 = 24) ways to rearrange the ups we picked, so we finally get:
displaystyle{frac{(10!/6!)}{4!} = frac{5040}{24} = 210 }
We're just picking the items to convert (10!/6!) and dividing out the redundancies (4!).
Source of answere :https://betterexplained.com/articles/navigate-a-grid-using-combinations-and-permutations/
answered 38 mins ago
PurplePurple
726113
$endgroup$
Instead of having 6 rights at 4 ups, imagine we start with 10 rights (r r r r r r r r r r).
Clearly this won't do: we need to change 4 of those rights into ups. How many ways can we pick 4 rights to change?
Well, we have 10 choices for the first 'right' to convert (see the combinations article). And 9 for the second, 8 for the third, and 7 choices for the final right-to-up conversion. There are 10 * 9 * 8 * 7 = 10!/6! = 5040 possibilities.
But, wait! We need to remove the redundancies: after all, converting moves #1 #2 #3 and #4 (in that order) is the same as converting #4 #3 #2 #1. We have 4! (4 * 3 * 2 * 1 = 24) ways to rearrange the ups we picked, so we finally get:
displaystyle{frac{(10!/6!)}{4!} = frac{5040}{24} = 210 }
We're just picking the items to convert (10!/6!) and dividing out the redundancies (4!).
Source of answere :https://betterexplained.com/articles/navigate-a-grid-using-combinations-and-permutations/
answered 38 mins ago
PurplePurple
726113
edited 59 secs ago
answered 38 mins ago
PurplePurple
726113
answered 38 mins ago
PurplePurple
726113
answered 38 mins ago
PurplePurple
726113
726113
1
$begingroup$
You should cite your source. This is clearly copied directly from the page you linked to in your comment above. Even the image is copied.
$endgroup$
– Daniel Mathias
14 mins ago
$begingroup$
@DanielMathias sorry sir I had made a mistake. I didn’t know that I should should provide link so a humble sorry . But the truth to be told mine answere was the same but lacked some explanation and a image , plz take back your devote . To support me as I am a beginner.
$endgroup$
– Purple
1 min ago
add a comment |
1
$begingroup$
You should cite your source. This is clearly copied directly from the page you linked to in your comment above. Even the image is copied.
$endgroup$
– Daniel Mathias
14 mins ago
$begingroup$
@DanielMathias sorry sir I had made a mistake. I didn’t know that I should should provide link so a humble sorry . But the truth to be told mine answere was the same but lacked some explanation and a image , plz take back your devote . To support me as I am a beginner.
$endgroup$
– Purple
1 min ago
1
1
$begingroup$
You should cite your source. This is clearly copied directly from the page you linked to in your comment above. Even the image is copied.
$endgroup$
– Daniel Mathias
14 mins ago
$begingroup$
You should cite your source. This is clearly copied directly from the page you linked to in your comment above. Even the image is copied.
$endgroup$
– Daniel Mathias
14 mins ago
$begingroup$
@DanielMathias sorry sir I had made a mistake. I didn’t know that I should should provide link so a humble sorry . But the truth to be told mine answere was the same but lacked some explanation and a image , plz take back your devote . To support me as I am a beginner.
$endgroup$
– Purple
1 min ago
$begingroup$
@DanielMathias sorry sir I had made a mistake. I didn’t know that I should should provide link so a humble sorry . But the truth to be told mine answere was the same but lacked some explanation and a image , plz take back your devote . To support me as I am a beginner.
$endgroup$
– Purple
1 min ago
add a comment |
$begingroup$
This is, I'm sure, answered somewhere else. It is also related to Pascal's triangle.
Simply fill out the grid as follows:
In this grid, each number represents the number of ways of getting to that particular intersection. And that number is precisely the number of ways to get to the intersection below it added to the number of ways to get to the intersection to the left of it.
answered 11 hours ago
Dr XorileDr Xorile
12.8k22569
$endgroup$
add a comment |
$begingroup$
This is, I'm sure, answered somewhere else. It is also related to Pascal's triangle.
Simply fill out the grid as follows:
In this grid, each number represents the number of ways of getting to that particular intersection. And that number is precisely the number of ways to get to the intersection below it added to the number of ways to get to the intersection to the left of it.
answered 11 hours ago
Dr XorileDr Xorile
12.8k22569
$endgroup$
add a comment |
$begingroup$
This is, I'm sure, answered somewhere else. It is also related to Pascal's triangle.
Simply fill out the grid as follows:
In this grid, each number represents the number of ways of getting to that particular intersection. And that number is precisely the number of ways to get to the intersection below it added to the number of ways to get to the intersection to the left of it.
answered 11 hours ago
Dr XorileDr Xorile
12.8k22569
$endgroup$
This is, I'm sure, answered somewhere else. It is also related to Pascal's triangle.
Simply fill out the grid as follows:
In this grid, each number represents the number of ways of getting to that particular intersection. And that number is precisely the number of ways to get to the intersection below it added to the number of ways to get to the intersection to the left of it.
answered 11 hours ago
Dr XorileDr Xorile
12.8k22569
edited 9 hours ago
answered 11 hours ago
Dr XorileDr Xorile
12.8k22569
answered 11 hours ago
Dr XorileDr Xorile
12.8k22569
answered 11 hours ago
Dr XorileDr Xorile
12.8k22569
12.8k22569
add a comment |
add a comment |
$begingroup$
A more mathematically oriented answer:
You have $10$ moves to make in total and you need to choose which $4$ of them are going to be up.
The number of ways to do that is $${10choose 4}=210$$
answered 4 hours ago
Arnaud MortierArnaud Mortier
688211
$endgroup$
add a comment |
$begingroup$
A more mathematically oriented answer:
You have $10$ moves to make in total and you need to choose which $4$ of them are going to be up.
The number of ways to do that is $${10choose 4}=210$$
answered 4 hours ago
Arnaud MortierArnaud Mortier
688211
$endgroup$
add a comment |
$begingroup$
A more mathematically oriented answer:
You have $10$ moves to make in total and you need to choose which $4$ of them are going to be up.
The number of ways to do that is $${10choose 4}=210$$
answered 4 hours ago
Arnaud MortierArnaud Mortier
688211
$endgroup$
A more mathematically oriented answer:
You have $10$ moves to make in total and you need to choose which $4$ of them are going to be up.
The number of ways to do that is $${10choose 4}=210$$
answered 4 hours ago
Arnaud MortierArnaud Mortier
688211
answered 4 hours ago
Arnaud MortierArnaud Mortier
688211
answered 4 hours ago
Arnaud MortierArnaud Mortier
688211
answered 4 hours ago
Arnaud MortierArnaud Mortier
688211
688211
add a comment |
add a comment |
$begingroup$
Seems complicated but , but I think this may help even though there are some good answer
There are ten ways to sum up .choose any four of them which are directing upwards
answered 1 hour ago
HavookHavook
212
New contributor
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Seems complicated but , but I think this may help even though there are some good answer
There are ten ways to sum up .choose any four of them which are directing upwards
answered 1 hour ago
HavookHavook
212
New contributor
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Seems complicated but , but I think this may help even though there are some good answer
There are ten ways to sum up .choose any four of them which are directing upwards
answered 1 hour ago
HavookHavook
212
New contributor
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Seems complicated but , but I think this may help even though there are some good answer
There are ten ways to sum up .choose any four of them which are directing upwards
answered 1 hour ago
HavookHavook
212
New contributor
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
HavookHavook
212
New contributor
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
HavookHavook
212
answered 1 hour ago
HavookHavook
212
212
New contributor
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Havook is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Write them out. Start with
rrrrrruuuu
rrrrruruuu
rrrrruuruu
...
uuuurrrrrr
and then count them up. You should have 210 in total.
Or:
$$binom30+binom41+binom52+binom63+binom74+binom85+binom96$$
$$=1+4+10+20+35+56+84$$
$$=210$$
answered 40 mins ago
JonMark PerryJonMark Perry
19.3k63991
$endgroup$
add a comment |
$begingroup$
Write them out. Start with
rrrrrruuuu
rrrrruruuu
rrrrruuruu
...
uuuurrrrrr
and then count them up. You should have 210 in total.
Or:
$$binom30+binom41+binom52+binom63+binom74+binom85+binom96$$
$$=1+4+10+20+35+56+84$$
$$=210$$
answered 40 mins ago
JonMark PerryJonMark Perry
19.3k63991
$endgroup$
add a comment |
$begingroup$
Write them out. Start with
rrrrrruuuu
rrrrruruuu
rrrrruuruu
...
uuuurrrrrr
and then count them up. You should have 210 in total.
Or:
$$binom30+binom41+binom52+binom63+binom74+binom85+binom96$$
$$=1+4+10+20+35+56+84$$
$$=210$$
answered 40 mins ago
JonMark PerryJonMark Perry
19.3k63991
$endgroup$
Write them out. Start with
rrrrrruuuu
rrrrruruuu
rrrrruuruu
...
uuuurrrrrr
and then count them up. You should have 210 in total.
Or:
$$binom30+binom41+binom52+binom63+binom74+binom85+binom96$$
$$=1+4+10+20+35+56+84$$
$$=210$$
answered 40 mins ago
JonMark PerryJonMark Perry
19.3k63991
answered 40 mins ago
JonMark PerryJonMark Perry
19.3k63991
answered 40 mins ago
JonMark PerryJonMark Perry
19.3k63991
answered 40 mins ago
JonMark PerryJonMark Perry
19.3k63991
19.3k63991
add a comment |
add a comment |
$begingroup$
YIPPY KAPPA IDDY GANGWA .........................................................................................................................☻☻☻☻☻☻☻☻☻☻☻☻☻☻
answered 1 hour ago
GangwaGangwa
1
New contributor
Gangwa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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YIPPY KAPPA IDDY GANGWA .........................................................................................................................☻☻☻☻☻☻☻☻☻☻☻☻☻☻
answered 1 hour ago
GangwaGangwa
1
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YIPPY KAPPA IDDY GANGWA .........................................................................................................................☻☻☻☻☻☻☻☻☻☻☻☻☻☻
answered 1 hour ago
GangwaGangwa
1
New contributor
Gangwa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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YIPPY KAPPA IDDY GANGWA .........................................................................................................................☻☻☻☻☻☻☻☻☻☻☻☻☻☻
answered 1 hour ago
GangwaGangwa
1
New contributor
Gangwa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
GangwaGangwa
1
New contributor
Gangwa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
GangwaGangwa
1
answered 1 hour ago
GangwaGangwa
1
1
New contributor
Gangwa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gangwa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gangwa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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I think its from this site betterexplained.com/articles/…
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– Purple
11 hours ago
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(In the future please be aware that for content you did not create yourself, proper attribution is required. You need to include (at minimum) where it came from—and any additional context you can provide is often helpful to solvers. Posts which use someone else's content without attribution are generally deleted.)
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– Rubio♦
9 hours ago