Street Combinatorics - 6 by 7 grid

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You go to school in a building located six blocks east and seven blocks north
of your home. So, in walking to school each day you go thirteen blocks. All streets in a rectangular pattern are available to you for walking.
In how many different paths can you go from home to school, walking only thirteen
blocks?

I want to say that the answer can be found knowing that there are $6!$ ways east and $7!$ ways north. Then, the answer would be $6!+ 7!$ .

I feel like this is way too simple of a solution to be correct.

combinatorics

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asked 1 hour ago

Ludwigthestud

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Ludwigthestud is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
3
down vote

favorite

1

You go to school in a building located six blocks east and seven blocks north
of your home. So, in walking to school each day you go thirteen blocks. All streets in a rectangular pattern are available to you for walking.
In how many different paths can you go from home to school, walking only thirteen
blocks?

I want to say that the answer can be found knowing that there are $6!$ ways east and $7!$ ways north. Then, the answer would be $6!+ 7!$ .

I feel like this is way too simple of a solution to be correct.

combinatorics

share|cite|improve this question

asked 1 hour ago

Ludwigthestud

262

New contributor

Ludwigthestud 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 | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

You go to school in a building located six blocks east and seven blocks north
of your home. So, in walking to school each day you go thirteen blocks. All streets in a rectangular pattern are available to you for walking.
In how many different paths can you go from home to school, walking only thirteen
blocks?

I want to say that the answer can be found knowing that there are $6!$ ways east and $7!$ ways north. Then, the answer would be $6!+ 7!$ .

I feel like this is way too simple of a solution to be correct.

combinatorics

share|cite|improve this question

asked 1 hour ago

Ludwigthestud

262

New contributor

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

You go to school in a building located six blocks east and seven blocks north
of your home. So, in walking to school each day you go thirteen blocks. All streets in a rectangular pattern are available to you for walking.
In how many different paths can you go from home to school, walking only thirteen
blocks?

I want to say that the answer can be found knowing that there are $6!$ ways east and $7!$ ways north. Then, the answer would be $6!+ 7!$ .

I feel like this is way too simple of a solution to be correct.

combinatorics

combinatorics

share|cite|improve this question

asked 1 hour ago

Ludwigthestud

262

New contributor

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

share|cite|improve this question

asked 1 hour ago

Ludwigthestud

262

New contributor

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

share|cite|improve this question

share|cite|improve this question

asked 1 hour ago

Ludwigthestud

262

New contributor

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

asked 1 hour ago

Ludwigthestud

262

asked 1 hour ago

Ludwigthestud

262

262

New contributor

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

New contributor

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

Ludwigthestud 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 | 

2 Answers
2

active

oldest

votes

up vote
4
down vote

Let's assume the only moves you are allowed are moving north and moving east. Denote a move north as n and a move east as e. Hence, we need to make 7 n moves and 6 e moves, and we seek to compute the number of arrangements of these moves.

This is equivalent to the problem

nnnnnnneeeeee

How many distinct rearrangements are there of the above letters. Through some combinatorics, we find the answer is $$13choose6=frac13!6!7!=colorred1716$$

share|cite|improve this answer

answered 1 hour ago

Rushabh Mehta

2,178221

add a comment | 

up vote
3
down vote

Consider this:

You have to go north 7 times and go east 6 times. How can you slip in the 6 "east moves" into 7 "north moves"?

The answer is then $binom6+76=binom136=1716$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

abc...

1,897527

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you messed up the combination notation...
  – Rushabh Mehta
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RushabhMehta I did. Thank you for point it out.
  – abc...
  1 hour ago
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

Let's assume the only moves you are allowed are moving north and moving east. Denote a move north as n and a move east as e. Hence, we need to make 7 n moves and 6 e moves, and we seek to compute the number of arrangements of these moves.

This is equivalent to the problem

nnnnnnneeeeee

How many distinct rearrangements are there of the above letters. Through some combinatorics, we find the answer is $$13choose6=frac13!6!7!=colorred1716$$

share|cite|improve this answer

answered 1 hour ago

Rushabh Mehta

2,178221

add a comment | 

up vote
4
down vote

Let's assume the only moves you are allowed are moving north and moving east. Denote a move north as n and a move east as e. Hence, we need to make 7 n moves and 6 e moves, and we seek to compute the number of arrangements of these moves.

This is equivalent to the problem

nnnnnnneeeeee

How many distinct rearrangements are there of the above letters. Through some combinatorics, we find the answer is $$13choose6=frac13!6!7!=colorred1716$$

share|cite|improve this answer

answered 1 hour ago

Rushabh Mehta

2,178221

add a comment | 

up vote
4
down vote

up vote
4
down vote

Let's assume the only moves you are allowed are moving north and moving east. Denote a move north as n and a move east as e. Hence, we need to make 7 n moves and 6 e moves, and we seek to compute the number of arrangements of these moves.

This is equivalent to the problem

nnnnnnneeeeee

How many distinct rearrangements are there of the above letters. Through some combinatorics, we find the answer is $$13choose6=frac13!6!7!=colorred1716$$

share|cite|improve this answer

answered 1 hour ago

Rushabh Mehta

2,178221

Let's assume the only moves you are allowed are moving north and moving east. Denote a move north as n and a move east as e. Hence, we need to make 7 n moves and 6 e moves, and we seek to compute the number of arrangements of these moves.

This is equivalent to the problem

nnnnnnneeeeee

How many distinct rearrangements are there of the above letters. Through some combinatorics, we find the answer is $$13choose6=frac13!6!7!=colorred1716$$

share|cite|improve this answer

answered 1 hour ago

Rushabh Mehta

2,178221

share|cite|improve this answer

share|cite|improve this answer

answered 1 hour ago

Rushabh Mehta

2,178221

answered 1 hour ago

Rushabh Mehta

2,178221

answered 1 hour ago

Rushabh Mehta

2,178221

2,178221

add a comment | 

add a comment | 

up vote
3
down vote

Consider this:

You have to go north 7 times and go east 6 times. How can you slip in the 6 "east moves" into 7 "north moves"?

The answer is then $binom6+76=binom136=1716$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

abc...

1,897527

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you messed up the combination notation...
  – Rushabh Mehta
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RushabhMehta I did. Thank you for point it out.
  – abc...
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

Consider this:

You have to go north 7 times and go east 6 times. How can you slip in the 6 "east moves" into 7 "north moves"?

The answer is then $binom6+76=binom136=1716$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

abc...

1,897527

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you messed up the combination notation...
  – Rushabh Mehta
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RushabhMehta I did. Thank you for point it out.
  – abc...
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Consider this:

You have to go north 7 times and go east 6 times. How can you slip in the 6 "east moves" into 7 "north moves"?

The answer is then $binom6+76=binom136=1716$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

abc...

1,897527

Consider this:

You have to go north 7 times and go east 6 times. How can you slip in the 6 "east moves" into 7 "north moves"?

The answer is then $binom6+76=binom136=1716$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

abc...

1,897527

share|cite|improve this answer

share|cite|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 1 hour ago

abc...

1,897527

answered 1 hour ago

abc...

1,897527

answered 1 hour ago

abc...

1,897527

1,897527

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you messed up the combination notation...
  – Rushabh Mehta
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RushabhMehta I did. Thank you for point it out.
  – abc...
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you messed up the combination notation...
  – Rushabh Mehta
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RushabhMehta I did. Thank you for point it out.
  – abc...
  1 hour ago
  

  
  
  
  

I think you messed up the combination notation...
– Rushabh Mehta
1 hour ago

I think you messed up the combination notation...
– Rushabh Mehta
1 hour ago

@RushabhMehta I did. Thank you for point it out.
– abc...
1 hour ago

@RushabhMehta I did. Thank you for point it out.
– abc...
1 hour ago

add a comment | 

Ludwigthestud is a new contributor. Be nice, and check out our Code of Conduct.

 

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