Mixing
Counting numbers of ways to reach from point A to B?
$begingroup$
Please check the attachment for the figure.
In how many ways, can a person go from point A to point B if he or she could only move down or right?
My answer is 9!/(5!4!).
Please let me know if I am right or wrong? And correct method too?
combinatorics
asked Jan 2 at 13:35
Amogh JoshiAmogh Joshi
164
$endgroup$
add a comment |
$begingroup$
Please check the attachment for the figure.
In how many ways, can a person go from point A to point B if he or she could only move down or right?
My answer is 9!/(5!4!).
Please let me know if I am right or wrong? And correct method too?
combinatorics
asked Jan 2 at 13:35
Amogh JoshiAmogh Joshi
164
$endgroup$
$begingroup$
My answer is 126 ways. Definitely more than 1 ways.
$endgroup$
– Amogh Joshi
Jan 2 at 13:42
$begingroup$
No. In a straight line, he could move either right or down. Don’t think of it as a whole block.
$endgroup$
– Amogh Joshi
Jan 2 at 13:48
add a comment |
$begingroup$
Please check the attachment for the figure.
In how many ways, can a person go from point A to point B if he or she could only move down or right?
My answer is 9!/(5!4!).
Please let me know if I am right or wrong? And correct method too?
combinatorics
asked Jan 2 at 13:35
Amogh JoshiAmogh Joshi
164
$endgroup$
Please check the attachment for the figure.
In how many ways, can a person go from point A to point B if he or she could only move down or right?
My answer is 9!/(5!4!).
Please let me know if I am right or wrong? And correct method too?
combinatorics
combinatorics
asked Jan 2 at 13:35
Amogh JoshiAmogh Joshi
164
asked Jan 2 at 13:35
Amogh JoshiAmogh Joshi
164
asked Jan 2 at 13:35
Amogh JoshiAmogh Joshi
164
asked Jan 2 at 13:35
Amogh JoshiAmogh Joshi
164
asked Jan 2 at 13:35
Amogh JoshiAmogh Joshi
164
164
$begingroup$
My answer is 126 ways. Definitely more than 1 ways.
$endgroup$
– Amogh Joshi
Jan 2 at 13:42
$begingroup$
No. In a straight line, he could move either right or down. Don’t think of it as a whole block.
$endgroup$
– Amogh Joshi
Jan 2 at 13:48
add a comment |
$begingroup$
My answer is 126 ways. Definitely more than 1 ways.
$endgroup$
– Amogh Joshi
Jan 2 at 13:42
$begingroup$
No. In a straight line, he could move either right or down. Don’t think of it as a whole block.
$endgroup$
– Amogh Joshi
Jan 2 at 13:48
$begingroup$
My answer is 126 ways. Definitely more than 1 ways.
$endgroup$
– Amogh Joshi
Jan 2 at 13:42
$begingroup$
My answer is 126 ways. Definitely more than 1 ways.
$endgroup$
– Amogh Joshi
Jan 2 at 13:42
$begingroup$
No. In a straight line, he could move either right or down. Don’t think of it as a whole block.
$endgroup$
– Amogh Joshi
Jan 2 at 13:48
$begingroup$
No. In a straight line, he could move either right or down. Don’t think of it as a whole block.
$endgroup$
– Amogh Joshi
Jan 2 at 13:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This diagram should make it clear that there are $29$ paths.
begin{array}{rrrrrrrrr}
text{A}&rightarrow&1&rightarrow&1\
downarrow&& downarrow&& downarrow\
1&rightarrow&2&rightarrow&3\
&& downarrow&& downarrow\
&&2&rightarrow&5\
&& downarrow&& downarrow\
&&2&rightarrow&7\
&& downarrow&& downarrow\
&&2&rightarrow&9&rightarrow&9&rightarrow&9\
&& downarrow&& downarrow&& downarrow&& downarrow\
&&2&rightarrow&11&rightarrow&20&rightarrow&text{B}
end{array}
answered Jan 2 at 16:17
Daniel MathiasDaniel Mathias
4586
$endgroup$
add a comment |
$begingroup$
How many ways from A to B in the figure below?
Now in your picture, are there more ways, fewer, or the same number of ways?
Does this picture have paths that yours does not?
Does your picture have paths that this one does not?
The general formula when you go from one corner of a rectangular grid to the diagonally opposite corner is $binom{W+H}{W} = frac{(W+H)!}{W!H!}.$ If you delete some of the lines in the grid, you lose the paths that go on those lines, and you need to do something more complicated to count the paths. Usually this will involve multiple cases, perhaps even inclusion-exclusion.
answered Jan 2 at 14:20
David KDavid K
53k340115
$endgroup$
$begingroup$
m.youtube.com/watch?v=f5AORqe1ADk
$endgroup$
– Amogh Joshi
Jan 2 at 14:23
$begingroup$
I know the answers to all my questions. Have you thought about what I said?
$endgroup$
– David K
Jan 2 at 14:25
$begingroup$
Based on this, 126 ways. Same number of ways. But my approch might be wrong to get 126 ways in my original question.
$endgroup$
– Amogh Joshi
Jan 2 at 14:26
$begingroup$
Your picture is different, but you say the number of paths is the same. How do you figure that?
$endgroup$
– David K
Jan 2 at 14:26
$begingroup$
I don’t know the number of ways for my original question. I was just curious to modify the question given in the video.
$endgroup$
– Amogh Joshi
Jan 2 at 14:28
|
show 4 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This diagram should make it clear that there are $29$ paths.
begin{array}{rrrrrrrrr}
text{A}&rightarrow&1&rightarrow&1\
downarrow&& downarrow&& downarrow\
1&rightarrow&2&rightarrow&3\
&& downarrow&& downarrow\
&&2&rightarrow&5\
&& downarrow&& downarrow\
&&2&rightarrow&7\
&& downarrow&& downarrow\
&&2&rightarrow&9&rightarrow&9&rightarrow&9\
&& downarrow&& downarrow&& downarrow&& downarrow\
&&2&rightarrow&11&rightarrow&20&rightarrow&text{B}
end{array}
answered Jan 2 at 16:17
Daniel MathiasDaniel Mathias
4586
$endgroup$
add a comment |
$begingroup$
This diagram should make it clear that there are $29$ paths.
begin{array}{rrrrrrrrr}
text{A}&rightarrow&1&rightarrow&1\
downarrow&& downarrow&& downarrow\
1&rightarrow&2&rightarrow&3\
&& downarrow&& downarrow\
&&2&rightarrow&5\
&& downarrow&& downarrow\
&&2&rightarrow&7\
&& downarrow&& downarrow\
&&2&rightarrow&9&rightarrow&9&rightarrow&9\
&& downarrow&& downarrow&& downarrow&& downarrow\
&&2&rightarrow&11&rightarrow&20&rightarrow&text{B}
end{array}
answered Jan 2 at 16:17
Daniel MathiasDaniel Mathias
4586
$endgroup$
add a comment |
$begingroup$
This diagram should make it clear that there are $29$ paths.
begin{array}{rrrrrrrrr}
text{A}&rightarrow&1&rightarrow&1\
downarrow&& downarrow&& downarrow\
1&rightarrow&2&rightarrow&3\
&& downarrow&& downarrow\
&&2&rightarrow&5\
&& downarrow&& downarrow\
&&2&rightarrow&7\
&& downarrow&& downarrow\
&&2&rightarrow&9&rightarrow&9&rightarrow&9\
&& downarrow&& downarrow&& downarrow&& downarrow\
&&2&rightarrow&11&rightarrow&20&rightarrow&text{B}
end{array}
answered Jan 2 at 16:17
Daniel MathiasDaniel Mathias
4586
$endgroup$
This diagram should make it clear that there are $29$ paths.
begin{array}{rrrrrrrrr}
text{A}&rightarrow&1&rightarrow&1\
downarrow&& downarrow&& downarrow\
1&rightarrow&2&rightarrow&3\
&& downarrow&& downarrow\
&&2&rightarrow&5\
&& downarrow&& downarrow\
&&2&rightarrow&7\
&& downarrow&& downarrow\
&&2&rightarrow&9&rightarrow&9&rightarrow&9\
&& downarrow&& downarrow&& downarrow&& downarrow\
&&2&rightarrow&11&rightarrow&20&rightarrow&text{B}
end{array}
answered Jan 2 at 16:17
Daniel MathiasDaniel Mathias
4586
answered Jan 2 at 16:17
Daniel MathiasDaniel Mathias
4586
answered Jan 2 at 16:17
Daniel MathiasDaniel Mathias
4586
answered Jan 2 at 16:17
Daniel MathiasDaniel Mathias
4586
4586
add a comment |
add a comment |
$begingroup$
How many ways from A to B in the figure below?
Now in your picture, are there more ways, fewer, or the same number of ways?
Does this picture have paths that yours does not?
Does your picture have paths that this one does not?
The general formula when you go from one corner of a rectangular grid to the diagonally opposite corner is $binom{W+H}{W} = frac{(W+H)!}{W!H!}.$ If you delete some of the lines in the grid, you lose the paths that go on those lines, and you need to do something more complicated to count the paths. Usually this will involve multiple cases, perhaps even inclusion-exclusion.
answered Jan 2 at 14:20
David KDavid K
53k340115
$endgroup$
$begingroup$
m.youtube.com/watch?v=f5AORqe1ADk
$endgroup$
– Amogh Joshi
Jan 2 at 14:23
$begingroup$
I know the answers to all my questions. Have you thought about what I said?
$endgroup$
– David K
Jan 2 at 14:25
$begingroup$
Based on this, 126 ways. Same number of ways. But my approch might be wrong to get 126 ways in my original question.
$endgroup$
– Amogh Joshi
Jan 2 at 14:26
$begingroup$
Your picture is different, but you say the number of paths is the same. How do you figure that?
$endgroup$
– David K
Jan 2 at 14:26
$begingroup$
I don’t know the number of ways for my original question. I was just curious to modify the question given in the video.
$endgroup$
– Amogh Joshi
Jan 2 at 14:28
|
show 4 more comments
$begingroup$
How many ways from A to B in the figure below?
Now in your picture, are there more ways, fewer, or the same number of ways?
Does this picture have paths that yours does not?
Does your picture have paths that this one does not?
The general formula when you go from one corner of a rectangular grid to the diagonally opposite corner is $binom{W+H}{W} = frac{(W+H)!}{W!H!}.$ If you delete some of the lines in the grid, you lose the paths that go on those lines, and you need to do something more complicated to count the paths. Usually this will involve multiple cases, perhaps even inclusion-exclusion.
answered Jan 2 at 14:20
David KDavid K
53k340115
$endgroup$
$begingroup$
m.youtube.com/watch?v=f5AORqe1ADk
$endgroup$
– Amogh Joshi
Jan 2 at 14:23
$begingroup$
I know the answers to all my questions. Have you thought about what I said?
$endgroup$
– David K
Jan 2 at 14:25
$begingroup$
Based on this, 126 ways. Same number of ways. But my approch might be wrong to get 126 ways in my original question.
$endgroup$
– Amogh Joshi
Jan 2 at 14:26
$begingroup$
Your picture is different, but you say the number of paths is the same. How do you figure that?
$endgroup$
– David K
Jan 2 at 14:26
$begingroup$
I don’t know the number of ways for my original question. I was just curious to modify the question given in the video.
$endgroup$
– Amogh Joshi
Jan 2 at 14:28
|
show 4 more comments
$begingroup$
How many ways from A to B in the figure below?
Now in your picture, are there more ways, fewer, or the same number of ways?
Does this picture have paths that yours does not?
Does your picture have paths that this one does not?
The general formula when you go from one corner of a rectangular grid to the diagonally opposite corner is $binom{W+H}{W} = frac{(W+H)!}{W!H!}.$ If you delete some of the lines in the grid, you lose the paths that go on those lines, and you need to do something more complicated to count the paths. Usually this will involve multiple cases, perhaps even inclusion-exclusion.
answered Jan 2 at 14:20
David KDavid K
53k340115
$endgroup$
How many ways from A to B in the figure below?
Now in your picture, are there more ways, fewer, or the same number of ways?
Does this picture have paths that yours does not?
Does your picture have paths that this one does not?
The general formula when you go from one corner of a rectangular grid to the diagonally opposite corner is $binom{W+H}{W} = frac{(W+H)!}{W!H!}.$ If you delete some of the lines in the grid, you lose the paths that go on those lines, and you need to do something more complicated to count the paths. Usually this will involve multiple cases, perhaps even inclusion-exclusion.
answered Jan 2 at 14:20
David KDavid K
53k340115
edited Jan 2 at 14:35
answered Jan 2 at 14:20
David KDavid K
53k340115
answered Jan 2 at 14:20
David KDavid K
53k340115
answered Jan 2 at 14:20
David KDavid K
53k340115
53k340115
$begingroup$
m.youtube.com/watch?v=f5AORqe1ADk
$endgroup$
– Amogh Joshi
Jan 2 at 14:23
$begingroup$
I know the answers to all my questions. Have you thought about what I said?
$endgroup$
– David K
Jan 2 at 14:25
$begingroup$
Based on this, 126 ways. Same number of ways. But my approch might be wrong to get 126 ways in my original question.
$endgroup$
– Amogh Joshi
Jan 2 at 14:26
$begingroup$
Your picture is different, but you say the number of paths is the same. How do you figure that?
$endgroup$
– David K
Jan 2 at 14:26
$begingroup$
I don’t know the number of ways for my original question. I was just curious to modify the question given in the video.
$endgroup$
– Amogh Joshi
Jan 2 at 14:28
|
show 4 more comments
$begingroup$
m.youtube.com/watch?v=f5AORqe1ADk
$endgroup$
– Amogh Joshi
Jan 2 at 14:23
$begingroup$
I know the answers to all my questions. Have you thought about what I said?
$endgroup$
– David K
Jan 2 at 14:25
$begingroup$
Based on this, 126 ways. Same number of ways. But my approch might be wrong to get 126 ways in my original question.
$endgroup$
– Amogh Joshi
Jan 2 at 14:26
$begingroup$
Your picture is different, but you say the number of paths is the same. How do you figure that?
$endgroup$
– David K
Jan 2 at 14:26
$begingroup$
I don’t know the number of ways for my original question. I was just curious to modify the question given in the video.
$endgroup$
– Amogh Joshi
Jan 2 at 14:28
$begingroup$
m.youtube.com/watch?v=f5AORqe1ADk
$endgroup$
– Amogh Joshi
Jan 2 at 14:23
$begingroup$
m.youtube.com/watch?v=f5AORqe1ADk
$endgroup$
– Amogh Joshi
Jan 2 at 14:23
$begingroup$
I know the answers to all my questions. Have you thought about what I said?
$endgroup$
– David K
Jan 2 at 14:25
$begingroup$
I know the answers to all my questions. Have you thought about what I said?
$endgroup$
– David K
Jan 2 at 14:25
$begingroup$
Based on this, 126 ways. Same number of ways. But my approch might be wrong to get 126 ways in my original question.
$endgroup$
– Amogh Joshi
Jan 2 at 14:26
$begingroup$
Based on this, 126 ways. Same number of ways. But my approch might be wrong to get 126 ways in my original question.
$endgroup$
– Amogh Joshi
Jan 2 at 14:26
$begingroup$
Your picture is different, but you say the number of paths is the same. How do you figure that?
$endgroup$
– David K
Jan 2 at 14:26
$begingroup$
Your picture is different, but you say the number of paths is the same. How do you figure that?
$endgroup$
– David K
Jan 2 at 14:26
$begingroup$
I don’t know the number of ways for my original question. I was just curious to modify the question given in the video.
$endgroup$
– Amogh Joshi
Jan 2 at 14:28
$begingroup$
I don’t know the number of ways for my original question. I was just curious to modify the question given in the video.
$endgroup$
– Amogh Joshi
Jan 2 at 14:28
|
show 4 more comments
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$begingroup$
My answer is 126 ways. Definitely more than 1 ways.
$endgroup$
– Amogh Joshi
Jan 2 at 13:42
$begingroup$
No. In a straight line, he could move either right or down. Don’t think of it as a whole block.
$endgroup$
– Amogh Joshi
Jan 2 at 13:48