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Number of ways to colour gridcells
The 2019 Stack Overflow Developer Survey Results Are InHow many ways to paint a rectangleNumber of $4times K$ board arrangements such that there exist no square of size 2 having all black cellsCalculating the number of all possible connected regions on a discrete gridNumber of ways of selecting cells from a gridNumber of $n^2times n^2$ permutation matrices with a 1 in each $ntimes n$ subgridIn how many ways can the four walls of a room be painted with three colours so that no two adjacent walls have the same colour?Number of ways to colour a $2×N$ gridcount the number of ways to partition a gridIn how many ways can we distribute 6 identical balls between 10 different cells, such that exactly one cell will contain exactly 3 balls?Different ways to count number of ways to colour a $4times4$ grid with 2 colours?
$begingroup$
Given a $6times 3$ grid, we would like to colour four cells with black in a way that exactly four of the columns contain one black square each.
How many ways are there to fulfill this colouring ?
My answer would be : $18 times 15 times 12 times 9 $
18 possibilities to colour the first cell,
15 to colour the second,
12 to colour the third,
9 to colour the fouth.
Is my answer correct ?
Thanks
combinatorics
asked Apr 6 at 21:56
ahmedahmed
474
$endgroup$
add a comment |
$begingroup$
Given a $6times 3$ grid, we would like to colour four cells with black in a way that exactly four of the columns contain one black square each.
How many ways are there to fulfill this colouring ?
My answer would be : $18 times 15 times 12 times 9 $
18 possibilities to colour the first cell,
15 to colour the second,
12 to colour the third,
9 to colour the fouth.
Is my answer correct ?
Thanks
combinatorics
asked Apr 6 at 21:56
ahmedahmed
474
$endgroup$
add a comment |
$begingroup$
Given a $6times 3$ grid, we would like to colour four cells with black in a way that exactly four of the columns contain one black square each.
How many ways are there to fulfill this colouring ?
My answer would be : $18 times 15 times 12 times 9 $
18 possibilities to colour the first cell,
15 to colour the second,
12 to colour the third,
9 to colour the fouth.
Is my answer correct ?
Thanks
combinatorics
asked Apr 6 at 21:56
ahmedahmed
474
$endgroup$
Given a $6times 3$ grid, we would like to colour four cells with black in a way that exactly four of the columns contain one black square each.
How many ways are there to fulfill this colouring ?
My answer would be : $18 times 15 times 12 times 9 $
18 possibilities to colour the first cell,
15 to colour the second,
12 to colour the third,
9 to colour the fouth.
Is my answer correct ?
Thanks
combinatorics
combinatorics
asked Apr 6 at 21:56
ahmedahmed
474
asked Apr 6 at 21:56
ahmedahmed
474
asked Apr 6 at 21:56
ahmedahmed
474
asked Apr 6 at 21:56
ahmedahmed
474
asked Apr 6 at 21:56
ahmedahmed
474
474
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your answer is almost correct; you have overcounted by the number of different orders in which you can pick the same black squares. You can pick the same $4$ black squares in $4!$ different ways, so you should divide your current answer by $24$.
Another way to see this, is that you can pick $4$ different columns out of $6$ for the black squares, and then pick $1$ row out of $3$ for each of the $4$ black squares. This gives you $binom64times3^4$.
answered Apr 6 at 22:07
ServaesServaes
30.3k342101
$endgroup$
2
$begingroup$
Unfortunately the OP's answer is not the same. Your solution is $1215$, while the OP's solution is $29160$
$endgroup$
– Haris Gusic
Apr 6 at 22:16
1
$begingroup$
@HarisGusic You are absolutely right, I have updated my answer.
$endgroup$
– Servaes
Apr 6 at 22:18
add a comment |
$begingroup$
No, your answer is not correct. Let's say that the first cells you color in order are $(1,1), (1,2), (1,3), (1,4)$. Then, in another attempt you can color them in another order, say in reverse, $(1,4), (1,3), (1,2), (1,1)$. See that these two approaches give you the same configuration. Therefore, by using your approach, you have counted some configurations twice.
The correct way to solve this problem is to first pick which columns are going to contain black cells. This can be done in $binom64$ ways. Then, you need to fill one cell of each of these rows, which can be done in $3^4$ ways. Applying the multiplicative rule, you get the result $binom64cdot 3^4$.
answered Apr 6 at 22:12
Haris GusicHaris Gusic
3,531627
$endgroup$
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
$begingroup$
Your answer is almost correct; you have overcounted by the number of different orders in which you can pick the same black squares. You can pick the same $4$ black squares in $4!$ different ways, so you should divide your current answer by $24$.
Another way to see this, is that you can pick $4$ different columns out of $6$ for the black squares, and then pick $1$ row out of $3$ for each of the $4$ black squares. This gives you $binom64times3^4$.
answered Apr 6 at 22:07
ServaesServaes
30.3k342101
$endgroup$
2
$begingroup$
Unfortunately the OP's answer is not the same. Your solution is $1215$, while the OP's solution is $29160$
$endgroup$
– Haris Gusic
Apr 6 at 22:16
1
$begingroup$
@HarisGusic You are absolutely right, I have updated my answer.
$endgroup$
– Servaes
Apr 6 at 22:18
add a comment |
$begingroup$
Your answer is almost correct; you have overcounted by the number of different orders in which you can pick the same black squares. You can pick the same $4$ black squares in $4!$ different ways, so you should divide your current answer by $24$.
Another way to see this, is that you can pick $4$ different columns out of $6$ for the black squares, and then pick $1$ row out of $3$ for each of the $4$ black squares. This gives you $binom64times3^4$.
answered Apr 6 at 22:07
ServaesServaes
30.3k342101
$endgroup$
2
$begingroup$
Unfortunately the OP's answer is not the same. Your solution is $1215$, while the OP's solution is $29160$
$endgroup$
– Haris Gusic
Apr 6 at 22:16
1
$begingroup$
@HarisGusic You are absolutely right, I have updated my answer.
$endgroup$
– Servaes
Apr 6 at 22:18
add a comment |
$begingroup$
Your answer is almost correct; you have overcounted by the number of different orders in which you can pick the same black squares. You can pick the same $4$ black squares in $4!$ different ways, so you should divide your current answer by $24$.
Another way to see this, is that you can pick $4$ different columns out of $6$ for the black squares, and then pick $1$ row out of $3$ for each of the $4$ black squares. This gives you $binom64times3^4$.
answered Apr 6 at 22:07
ServaesServaes
30.3k342101
$endgroup$
Your answer is almost correct; you have overcounted by the number of different orders in which you can pick the same black squares. You can pick the same $4$ black squares in $4!$ different ways, so you should divide your current answer by $24$.
Another way to see this, is that you can pick $4$ different columns out of $6$ for the black squares, and then pick $1$ row out of $3$ for each of the $4$ black squares. This gives you $binom64times3^4$.
answered Apr 6 at 22:07
ServaesServaes
30.3k342101
edited Apr 6 at 22:18
answered Apr 6 at 22:07
ServaesServaes
30.3k342101
answered Apr 6 at 22:07
ServaesServaes
30.3k342101
answered Apr 6 at 22:07
ServaesServaes
30.3k342101
30.3k342101
2
$begingroup$
Unfortunately the OP's answer is not the same. Your solution is $1215$, while the OP's solution is $29160$
$endgroup$
– Haris Gusic
Apr 6 at 22:16
1
$begingroup$
@HarisGusic You are absolutely right, I have updated my answer.
$endgroup$
– Servaes
Apr 6 at 22:18
add a comment |
2
$begingroup$
Unfortunately the OP's answer is not the same. Your solution is $1215$, while the OP's solution is $29160$
$endgroup$
– Haris Gusic
Apr 6 at 22:16
1
$begingroup$
@HarisGusic You are absolutely right, I have updated my answer.
$endgroup$
– Servaes
Apr 6 at 22:18
2
2
$begingroup$
Unfortunately the OP's answer is not the same. Your solution is $1215$, while the OP's solution is $29160$
$endgroup$
– Haris Gusic
Apr 6 at 22:16
$begingroup$
Unfortunately the OP's answer is not the same. Your solution is $1215$, while the OP's solution is $29160$
$endgroup$
– Haris Gusic
Apr 6 at 22:16
1
1
$begingroup$
@HarisGusic You are absolutely right, I have updated my answer.
$endgroup$
– Servaes
Apr 6 at 22:18
$begingroup$
@HarisGusic You are absolutely right, I have updated my answer.
$endgroup$
– Servaes
Apr 6 at 22:18
add a comment |
$begingroup$
No, your answer is not correct. Let's say that the first cells you color in order are $(1,1), (1,2), (1,3), (1,4)$. Then, in another attempt you can color them in another order, say in reverse, $(1,4), (1,3), (1,2), (1,1)$. See that these two approaches give you the same configuration. Therefore, by using your approach, you have counted some configurations twice.
The correct way to solve this problem is to first pick which columns are going to contain black cells. This can be done in $binom64$ ways. Then, you need to fill one cell of each of these rows, which can be done in $3^4$ ways. Applying the multiplicative rule, you get the result $binom64cdot 3^4$.
answered Apr 6 at 22:12
Haris GusicHaris Gusic
3,531627
$endgroup$
add a comment |
$begingroup$
No, your answer is not correct. Let's say that the first cells you color in order are $(1,1), (1,2), (1,3), (1,4)$. Then, in another attempt you can color them in another order, say in reverse, $(1,4), (1,3), (1,2), (1,1)$. See that these two approaches give you the same configuration. Therefore, by using your approach, you have counted some configurations twice.
The correct way to solve this problem is to first pick which columns are going to contain black cells. This can be done in $binom64$ ways. Then, you need to fill one cell of each of these rows, which can be done in $3^4$ ways. Applying the multiplicative rule, you get the result $binom64cdot 3^4$.
answered Apr 6 at 22:12
Haris GusicHaris Gusic
3,531627
$endgroup$
add a comment |
$begingroup$
No, your answer is not correct. Let's say that the first cells you color in order are $(1,1), (1,2), (1,3), (1,4)$. Then, in another attempt you can color them in another order, say in reverse, $(1,4), (1,3), (1,2), (1,1)$. See that these two approaches give you the same configuration. Therefore, by using your approach, you have counted some configurations twice.
The correct way to solve this problem is to first pick which columns are going to contain black cells. This can be done in $binom64$ ways. Then, you need to fill one cell of each of these rows, which can be done in $3^4$ ways. Applying the multiplicative rule, you get the result $binom64cdot 3^4$.
answered Apr 6 at 22:12
Haris GusicHaris Gusic
3,531627
$endgroup$
No, your answer is not correct. Let's say that the first cells you color in order are $(1,1), (1,2), (1,3), (1,4)$. Then, in another attempt you can color them in another order, say in reverse, $(1,4), (1,3), (1,2), (1,1)$. See that these two approaches give you the same configuration. Therefore, by using your approach, you have counted some configurations twice.
The correct way to solve this problem is to first pick which columns are going to contain black cells. This can be done in $binom64$ ways. Then, you need to fill one cell of each of these rows, which can be done in $3^4$ ways. Applying the multiplicative rule, you get the result $binom64cdot 3^4$.
answered Apr 6 at 22:12
Haris GusicHaris Gusic
3,531627
answered Apr 6 at 22:12
Haris GusicHaris Gusic
3,531627
answered Apr 6 at 22:12
Haris GusicHaris Gusic
3,531627
answered Apr 6 at 22:12
Haris GusicHaris Gusic
3,531627
3,531627
add a comment |
add a comment |
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