Counters on a 5 by 5 grid The 2019 Stack Overflow Developer Survey Results Are InPlacing 5 pieces on a 5x5 grid with no main diagonalColor an $ntimes n$ square with $n$ colorsNumber of $1$s in a binary gridProblem regarding filling squares inside a $ntimes n$ grid.Filling a grid so as not to completely fill any row, column or *any* diagonalFilling a 40 x 40 grid with 3x3 squaresPlacing 5 pieces on a 5x5 grid with no main diagonalNumber of ways of selecting cells from a gridCounters on a Chessboard (BMO 2010/11)Placing counters on a 5x5 gridChessboard Combinatorial Problem
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Counters on a 5 by 5 grid
The 2019 Stack Overflow Developer Survey Results Are InPlacing 5 pieces on a 5x5 grid with no main diagonalColor an $ntimes n$ square with $n$ colorsNumber of $1$s in a binary gridProblem regarding filling squares inside a $ntimes n$ grid.Filling a grid so as not to completely fill any row, column or *any* diagonalFilling a 40 x 40 grid with 3x3 squaresPlacing 5 pieces on a 5x5 grid with no main diagonalNumber of ways of selecting cells from a gridCounters on a Chessboard (BMO 2010/11)Placing counters on a 5x5 gridChessboard Combinatorial Problem
$begingroup$
Given a $5times5$ grid, find the number of ways of placing 5 counters in the squares, so that each row and each column contains exactly one counter. One possible way is 
I know that there are 44 ways without the main diagonal from this link Placing 5 pieces on a 5x5 grid with no main diagonal assuming that answer is correct, but I don't know ho to find the number of ways with no missing main diagonals. Can someone help me?
combinatorics
asked Mar 23 at 3:11
sumisumi
634
$endgroup$
add a comment |
$begingroup$
Given a $5times5$ grid, find the number of ways of placing 5 counters in the squares, so that each row and each column contains exactly one counter. One possible way is
I know that there are 44 ways without the main diagonal from this link Placing 5 pieces on a 5x5 grid with no main diagonal assuming that answer is correct, but I don't know ho to find the number of ways with no missing main diagonals. Can someone help me?
combinatorics
asked Mar 23 at 3:11
sumisumi
634
$endgroup$
add a comment |
$begingroup$
Given a $5times5$ grid, find the number of ways of placing 5 counters in the squares, so that each row and each column contains exactly one counter. One possible way is
I know that there are 44 ways without the main diagonal from this link Placing 5 pieces on a 5x5 grid with no main diagonal assuming that answer is correct, but I don't know ho to find the number of ways with no missing main diagonals. Can someone help me?
combinatorics
asked Mar 23 at 3:11
sumisumi
634
$endgroup$
Given a $5times5$ grid, find the number of ways of placing 5 counters in the squares, so that each row and each column contains exactly one counter. One possible way is
I know that there are 44 ways without the main diagonal from this link Placing 5 pieces on a 5x5 grid with no main diagonal assuming that answer is correct, but I don't know ho to find the number of ways with no missing main diagonals. Can someone help me?
combinatorics
combinatorics
asked Mar 23 at 3:11
sumisumi
634
asked Mar 23 at 3:11
sumisumi
634
asked Mar 23 at 3:11
sumisumi
634
asked Mar 23 at 3:11
sumisumi
634
asked Mar 23 at 3:11
sumisumi
634
634
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have $25$ choices for the first counter, $16$ for the second, $9$ for the third, $4$ for the fourth, and $1$ for the fifth. That is a total of $25cdot16cdot9cdot4cdot1=14400$. Since the counters are indistinguishable, you have to divide this number by $5!=120$. This yields $14400/120=120$.
answered Mar 23 at 3:17
JosuéJosué
3,51242672
$endgroup$
$begingroup$
Or ... it's just 5!
$endgroup$
– Don Thousand
Mar 23 at 3:20
add a comment |
$begingroup$
Sort the locations by row number, then any permutation of the column numbers is acceptable, so there are $5!=120$ arrangements.
answered Mar 23 at 3:16
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You have $25$ choices for the first counter, $16$ for the second, $9$ for the third, $4$ for the fourth, and $1$ for the fifth. That is a total of $25cdot16cdot9cdot4cdot1=14400$. Since the counters are indistinguishable, you have to divide this number by $5!=120$. This yields $14400/120=120$.
answered Mar 23 at 3:17
JosuéJosué
3,51242672
$endgroup$
$begingroup$
Or ... it's just 5!
$endgroup$
– Don Thousand
Mar 23 at 3:20
add a comment |
$begingroup$
You have $25$ choices for the first counter, $16$ for the second, $9$ for the third, $4$ for the fourth, and $1$ for the fifth. That is a total of $25cdot16cdot9cdot4cdot1=14400$. Since the counters are indistinguishable, you have to divide this number by $5!=120$. This yields $14400/120=120$.
answered Mar 23 at 3:17
JosuéJosué
3,51242672
$endgroup$
$begingroup$
Or ... it's just 5!
$endgroup$
– Don Thousand
Mar 23 at 3:20
add a comment |
$begingroup$
You have $25$ choices for the first counter, $16$ for the second, $9$ for the third, $4$ for the fourth, and $1$ for the fifth. That is a total of $25cdot16cdot9cdot4cdot1=14400$. Since the counters are indistinguishable, you have to divide this number by $5!=120$. This yields $14400/120=120$.
answered Mar 23 at 3:17
JosuéJosué
3,51242672
$endgroup$
You have $25$ choices for the first counter, $16$ for the second, $9$ for the third, $4$ for the fourth, and $1$ for the fifth. That is a total of $25cdot16cdot9cdot4cdot1=14400$. Since the counters are indistinguishable, you have to divide this number by $5!=120$. This yields $14400/120=120$.
answered Mar 23 at 3:17
JosuéJosué
3,51242672
answered Mar 23 at 3:17
JosuéJosué
3,51242672
answered Mar 23 at 3:17
JosuéJosué
3,51242672
answered Mar 23 at 3:17
JosuéJosué
3,51242672
3,51242672
$begingroup$
Or ... it's just 5!
$endgroup$
– Don Thousand
Mar 23 at 3:20
add a comment |
$begingroup$
Or ... it's just 5!
$endgroup$
– Don Thousand
Mar 23 at 3:20
$begingroup$
Or ... it's just 5!
$endgroup$
– Don Thousand
Mar 23 at 3:20
$begingroup$
Or ... it's just 5!
$endgroup$
– Don Thousand
Mar 23 at 3:20
add a comment |
$begingroup$
Sort the locations by row number, then any permutation of the column numbers is acceptable, so there are $5!=120$ arrangements.
answered Mar 23 at 3:16
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Sort the locations by row number, then any permutation of the column numbers is acceptable, so there are $5!=120$ arrangements.
answered Mar 23 at 3:16
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Sort the locations by row number, then any permutation of the column numbers is acceptable, so there are $5!=120$ arrangements.
answered Mar 23 at 3:16
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Sort the locations by row number, then any permutation of the column numbers is acceptable, so there are $5!=120$ arrangements.
answered Mar 23 at 3:16
Ross MillikanRoss Millikan
301k24200375
answered Mar 23 at 3:16
Ross MillikanRoss Millikan
301k24200375
answered Mar 23 at 3:16
Ross MillikanRoss Millikan
301k24200375
answered Mar 23 at 3:16
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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