Number of ways to arrange $a,a,b,b,c,d$ in the grid such that no row is empty

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Find Number of ways to arrange $a,a,b,b,c,d$ in the grid below such that no row is empty



enter image description here



I tried to use principle of inclusion exclusion taking total possible minus empty rows.



that is we have required answer as



$$S=frac11!2!2!-left(R_1e+R_2e+R_3e+R_4e+R_5eright)+6 left(R_12eright)-2left(R_124eright)+R_24e$$



where suffix $e$ stands for empty



Then we get



$$S=frac11!4-6 left(frac3 times 8!4+frac2 times 10!4right)+6 frac7!4-2 frac6!4+frac9!4$$



But the answer is $14580$




combinatorics permutations inclusion-exclusion




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  • Good approach! What result did you get?
    – Cameron Buie
    Aug 3 at 2:28














up vote
1
down vote

favorite
1












Find Number of ways to arrange $a,a,b,b,c,d$ in the grid below such that no row is empty



enter image description here



I tried to use principle of inclusion exclusion taking total possible minus empty rows.



that is we have required answer as



$$S=frac11!2!2!-left(R_1e+R_2e+R_3e+R_4e+R_5eright)+6 left(R_12eright)-2left(R_124eright)+R_24e$$



where suffix $e$ stands for empty



Then we get



$$S=frac11!4-6 left(frac3 times 8!4+frac2 times 10!4right)+6 frac7!4-2 frac6!4+frac9!4$$



But the answer is $14580$




combinatorics permutations inclusion-exclusion




share|cite|improve this question





















  • Good approach! What result did you get?
    – Cameron Buie
    Aug 3 at 2:28












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Find Number of ways to arrange $a,a,b,b,c,d$ in the grid below such that no row is empty



enter image description here



I tried to use principle of inclusion exclusion taking total possible minus empty rows.



that is we have required answer as



$$S=frac11!2!2!-left(R_1e+R_2e+R_3e+R_4e+R_5eright)+6 left(R_12eright)-2left(R_124eright)+R_24e$$



where suffix $e$ stands for empty



Then we get



$$S=frac11!4-6 left(frac3 times 8!4+frac2 times 10!4right)+6 frac7!4-2 frac6!4+frac9!4$$



But the answer is $14580$




combinatorics permutations inclusion-exclusion




share|cite|improve this question













Find Number of ways to arrange $a,a,b,b,c,d$ in the grid below such that no row is empty



enter image description here



I tried to use principle of inclusion exclusion taking total possible minus empty rows.



that is we have required answer as



$$S=frac11!2!2!-left(R_1e+R_2e+R_3e+R_4e+R_5eright)+6 left(R_12eright)-2left(R_124eright)+R_24e$$



where suffix $e$ stands for empty



Then we get



$$S=frac11!4-6 left(frac3 times 8!4+frac2 times 10!4right)+6 frac7!4-2 frac6!4+frac9!4$$



But the answer is $14580$





combinatorics permutations inclusion-exclusion





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edited Aug 3 at 3:52









N. F. Taussig

37.9k93053




37.9k93053









asked Aug 3 at 2:26









Umesh shankar

2,19811018




2,19811018











  • Good approach! What result did you get?
    – Cameron Buie
    Aug 3 at 2:28
















  • Good approach! What result did you get?
    – Cameron Buie
    Aug 3 at 2:28















Good approach! What result did you get?
– Cameron Buie
Aug 3 at 2:28




Good approach! What result did you get?
– Cameron Buie
Aug 3 at 2:28










1 Answer
1






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oldest

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up vote
2
down vote



accepted










You have not accounted for the empty spaces in the grid.



There are $binom112$ ways to choose two of the eleven positions for the $a$s, $binom92$ ways to choose two of the remaining nine positions for the $b$s, $binom71$ ways to choose one of the remaining positions for the $c$, and $binom61$ ways to choose one of the remaining six positions for the $d$. Hence, you should have obtained
$$binom112binom92binom71binom61 = frac11!2!9! cdot frac9!2!7! cdot frac7!1!6! cdot frac6!1!5! = frac11!2!2!1!1!5!$$
distinguishable arrangements of two $a$s, two $b$s, one $c$, one $d$ in the grid without restriction. The factors of $2!$, $2!$, $1!$, $1!$, and $5!$ in the denominator account, respectively for the number of ways two $a$s, two $b$s, one $c$, one $d$, and five empty squares can be permuted among themselves without producing an arrangement that is distinguishable from the given arrangement. To see why, think of the problem as arranging two $a$s, two $b$s, one $c$, one $d$, and five $e$s in the eleven squares of the grid, where $e$ represents an empty square.



One empty row: The row can have either three squares or one square.



A row with three squares is empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom82$ ways to select two of the remaining eight positions for the $a$s, $binom62$ ways to choose two of the remaining six positions for the $b$s, $binom41$ ways to choose one of the remaining four positions for the $c$, and $binom31$ ways to choose one the remaining three positions for the $d$. Hence, there are
$$binom31binom82binom62binom41binom31$$
such arrangements.



A row with one square is empty: There are $binom21$ ways to choose which of the two rows with one square is empty, $binom102$ ways to select two of the remaining ten positions for the $a$s, $binom82$ ways to choose two of the remaining eight positions for the $b$s, $binom61$ ways to choose one of the remaining six positions for the $c$, and $binom51$ ways to choose one the remaining five positions for the $d$. Hence, there are
$$binom21binom102binom82binom61binom51$$
such arrangements.



Two empty rows: If two rows with three squares were empty, then there would not be a sufficient number of squares left in the grid for the six letters. Hence, we are left with two possibilities. Either a row with three squares and a row with one square are empty or both rows with one square are empty.



A row with three squares and a row with one square are empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom21$ ways to choose which of the rows with two square are empty, $binom72$ ways to select two of the remaining seven positions for the $a$s, $binom52$ ways to choose two of the remaining five positions for the $b$s, $binom31$ ways to choose one of the remaining three positions for the $c$, and $binom21$ ways to choose one the remaining two positions for the $d$. Hence, there are
$$binom31binom21binom72binom52binom31binom21$$
such arrangements.



Both rows with one square are empty: There are $binom92$ ways to select two of the remaining nine positions for the $a$s, $binom72$ ways to choose two of the remaining seven positions for the $b$s, $binom51$ ways to choose one of the remaining five positions for the $c$, and $binom41$ ways to choose one the remaining four positions for the $d$. Hence, there are
$$binom92binom72binom51binom41$$
such arrangements.



Three empty rows: Since at most five squares can be left empty, the only way this can occur is if both of the rows with one square and one of the rows with three square are empty.



There are $binom31$ ways to select which of the rows with three squares is empty, $binom62$ ways to select two of the remaining six positions for the $a$s, $binom42$ ways to choose two of the remaining four positions for the $b$s, $binom21$ ways to choose one of the remaining two positions for the $c$, and $binom11$ ways to choose the remaining position for the $d$. Hence, there are
$$binom31binom62binom42binom21binom11$$
such arrangements.



It is not possible to have more than three empty rows. By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$binom112binom92binom71binom61 - binom31binom82binom62binom41binom31 - binom21binom102binom82binom61binom51 + binom31binom21binom72binom52binom31binom21 + binom92binom72binom51binom41 - binom31binom62binom42binom21binom11$$






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    Good one thanks a ton
    – Umesh shankar
    2 days ago










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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










You have not accounted for the empty spaces in the grid.



There are $binom112$ ways to choose two of the eleven positions for the $a$s, $binom92$ ways to choose two of the remaining nine positions for the $b$s, $binom71$ ways to choose one of the remaining positions for the $c$, and $binom61$ ways to choose one of the remaining six positions for the $d$. Hence, you should have obtained
$$binom112binom92binom71binom61 = frac11!2!9! cdot frac9!2!7! cdot frac7!1!6! cdot frac6!1!5! = frac11!2!2!1!1!5!$$
distinguishable arrangements of two $a$s, two $b$s, one $c$, one $d$ in the grid without restriction. The factors of $2!$, $2!$, $1!$, $1!$, and $5!$ in the denominator account, respectively for the number of ways two $a$s, two $b$s, one $c$, one $d$, and five empty squares can be permuted among themselves without producing an arrangement that is distinguishable from the given arrangement. To see why, think of the problem as arranging two $a$s, two $b$s, one $c$, one $d$, and five $e$s in the eleven squares of the grid, where $e$ represents an empty square.



One empty row: The row can have either three squares or one square.



A row with three squares is empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom82$ ways to select two of the remaining eight positions for the $a$s, $binom62$ ways to choose two of the remaining six positions for the $b$s, $binom41$ ways to choose one of the remaining four positions for the $c$, and $binom31$ ways to choose one the remaining three positions for the $d$. Hence, there are
$$binom31binom82binom62binom41binom31$$
such arrangements.



A row with one square is empty: There are $binom21$ ways to choose which of the two rows with one square is empty, $binom102$ ways to select two of the remaining ten positions for the $a$s, $binom82$ ways to choose two of the remaining eight positions for the $b$s, $binom61$ ways to choose one of the remaining six positions for the $c$, and $binom51$ ways to choose one the remaining five positions for the $d$. Hence, there are
$$binom21binom102binom82binom61binom51$$
such arrangements.



Two empty rows: If two rows with three squares were empty, then there would not be a sufficient number of squares left in the grid for the six letters. Hence, we are left with two possibilities. Either a row with three squares and a row with one square are empty or both rows with one square are empty.



A row with three squares and a row with one square are empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom21$ ways to choose which of the rows with two square are empty, $binom72$ ways to select two of the remaining seven positions for the $a$s, $binom52$ ways to choose two of the remaining five positions for the $b$s, $binom31$ ways to choose one of the remaining three positions for the $c$, and $binom21$ ways to choose one the remaining two positions for the $d$. Hence, there are
$$binom31binom21binom72binom52binom31binom21$$
such arrangements.



Both rows with one square are empty: There are $binom92$ ways to select two of the remaining nine positions for the $a$s, $binom72$ ways to choose two of the remaining seven positions for the $b$s, $binom51$ ways to choose one of the remaining five positions for the $c$, and $binom41$ ways to choose one the remaining four positions for the $d$. Hence, there are
$$binom92binom72binom51binom41$$
such arrangements.



Three empty rows: Since at most five squares can be left empty, the only way this can occur is if both of the rows with one square and one of the rows with three square are empty.



There are $binom31$ ways to select which of the rows with three squares is empty, $binom62$ ways to select two of the remaining six positions for the $a$s, $binom42$ ways to choose two of the remaining four positions for the $b$s, $binom21$ ways to choose one of the remaining two positions for the $c$, and $binom11$ ways to choose the remaining position for the $d$. Hence, there are
$$binom31binom62binom42binom21binom11$$
such arrangements.



It is not possible to have more than three empty rows. By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$binom112binom92binom71binom61 - binom31binom82binom62binom41binom31 - binom21binom102binom82binom61binom51 + binom31binom21binom72binom52binom31binom21 + binom92binom72binom51binom41 - binom31binom62binom42binom21binom11$$






share|cite|improve this answer



















  • 1




    Good one thanks a ton
    – Umesh shankar
    2 days ago














up vote
2
down vote



accepted










You have not accounted for the empty spaces in the grid.



There are $binom112$ ways to choose two of the eleven positions for the $a$s, $binom92$ ways to choose two of the remaining nine positions for the $b$s, $binom71$ ways to choose one of the remaining positions for the $c$, and $binom61$ ways to choose one of the remaining six positions for the $d$. Hence, you should have obtained
$$binom112binom92binom71binom61 = frac11!2!9! cdot frac9!2!7! cdot frac7!1!6! cdot frac6!1!5! = frac11!2!2!1!1!5!$$
distinguishable arrangements of two $a$s, two $b$s, one $c$, one $d$ in the grid without restriction. The factors of $2!$, $2!$, $1!$, $1!$, and $5!$ in the denominator account, respectively for the number of ways two $a$s, two $b$s, one $c$, one $d$, and five empty squares can be permuted among themselves without producing an arrangement that is distinguishable from the given arrangement. To see why, think of the problem as arranging two $a$s, two $b$s, one $c$, one $d$, and five $e$s in the eleven squares of the grid, where $e$ represents an empty square.



One empty row: The row can have either three squares or one square.



A row with three squares is empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom82$ ways to select two of the remaining eight positions for the $a$s, $binom62$ ways to choose two of the remaining six positions for the $b$s, $binom41$ ways to choose one of the remaining four positions for the $c$, and $binom31$ ways to choose one the remaining three positions for the $d$. Hence, there are
$$binom31binom82binom62binom41binom31$$
such arrangements.



A row with one square is empty: There are $binom21$ ways to choose which of the two rows with one square is empty, $binom102$ ways to select two of the remaining ten positions for the $a$s, $binom82$ ways to choose two of the remaining eight positions for the $b$s, $binom61$ ways to choose one of the remaining six positions for the $c$, and $binom51$ ways to choose one the remaining five positions for the $d$. Hence, there are
$$binom21binom102binom82binom61binom51$$
such arrangements.



Two empty rows: If two rows with three squares were empty, then there would not be a sufficient number of squares left in the grid for the six letters. Hence, we are left with two possibilities. Either a row with three squares and a row with one square are empty or both rows with one square are empty.



A row with three squares and a row with one square are empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom21$ ways to choose which of the rows with two square are empty, $binom72$ ways to select two of the remaining seven positions for the $a$s, $binom52$ ways to choose two of the remaining five positions for the $b$s, $binom31$ ways to choose one of the remaining three positions for the $c$, and $binom21$ ways to choose one the remaining two positions for the $d$. Hence, there are
$$binom31binom21binom72binom52binom31binom21$$
such arrangements.



Both rows with one square are empty: There are $binom92$ ways to select two of the remaining nine positions for the $a$s, $binom72$ ways to choose two of the remaining seven positions for the $b$s, $binom51$ ways to choose one of the remaining five positions for the $c$, and $binom41$ ways to choose one the remaining four positions for the $d$. Hence, there are
$$binom92binom72binom51binom41$$
such arrangements.



Three empty rows: Since at most five squares can be left empty, the only way this can occur is if both of the rows with one square and one of the rows with three square are empty.



There are $binom31$ ways to select which of the rows with three squares is empty, $binom62$ ways to select two of the remaining six positions for the $a$s, $binom42$ ways to choose two of the remaining four positions for the $b$s, $binom21$ ways to choose one of the remaining two positions for the $c$, and $binom11$ ways to choose the remaining position for the $d$. Hence, there are
$$binom31binom62binom42binom21binom11$$
such arrangements.



It is not possible to have more than three empty rows. By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$binom112binom92binom71binom61 - binom31binom82binom62binom41binom31 - binom21binom102binom82binom61binom51 + binom31binom21binom72binom52binom31binom21 + binom92binom72binom51binom41 - binom31binom62binom42binom21binom11$$






share|cite|improve this answer



















  • 1




    Good one thanks a ton
    – Umesh shankar
    2 days ago












up vote
2
down vote



accepted







up vote
2
down vote



accepted






You have not accounted for the empty spaces in the grid.



There are $binom112$ ways to choose two of the eleven positions for the $a$s, $binom92$ ways to choose two of the remaining nine positions for the $b$s, $binom71$ ways to choose one of the remaining positions for the $c$, and $binom61$ ways to choose one of the remaining six positions for the $d$. Hence, you should have obtained
$$binom112binom92binom71binom61 = frac11!2!9! cdot frac9!2!7! cdot frac7!1!6! cdot frac6!1!5! = frac11!2!2!1!1!5!$$
distinguishable arrangements of two $a$s, two $b$s, one $c$, one $d$ in the grid without restriction. The factors of $2!$, $2!$, $1!$, $1!$, and $5!$ in the denominator account, respectively for the number of ways two $a$s, two $b$s, one $c$, one $d$, and five empty squares can be permuted among themselves without producing an arrangement that is distinguishable from the given arrangement. To see why, think of the problem as arranging two $a$s, two $b$s, one $c$, one $d$, and five $e$s in the eleven squares of the grid, where $e$ represents an empty square.



One empty row: The row can have either three squares or one square.



A row with three squares is empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom82$ ways to select two of the remaining eight positions for the $a$s, $binom62$ ways to choose two of the remaining six positions for the $b$s, $binom41$ ways to choose one of the remaining four positions for the $c$, and $binom31$ ways to choose one the remaining three positions for the $d$. Hence, there are
$$binom31binom82binom62binom41binom31$$
such arrangements.



A row with one square is empty: There are $binom21$ ways to choose which of the two rows with one square is empty, $binom102$ ways to select two of the remaining ten positions for the $a$s, $binom82$ ways to choose two of the remaining eight positions for the $b$s, $binom61$ ways to choose one of the remaining six positions for the $c$, and $binom51$ ways to choose one the remaining five positions for the $d$. Hence, there are
$$binom21binom102binom82binom61binom51$$
such arrangements.



Two empty rows: If two rows with three squares were empty, then there would not be a sufficient number of squares left in the grid for the six letters. Hence, we are left with two possibilities. Either a row with three squares and a row with one square are empty or both rows with one square are empty.



A row with three squares and a row with one square are empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom21$ ways to choose which of the rows with two square are empty, $binom72$ ways to select two of the remaining seven positions for the $a$s, $binom52$ ways to choose two of the remaining five positions for the $b$s, $binom31$ ways to choose one of the remaining three positions for the $c$, and $binom21$ ways to choose one the remaining two positions for the $d$. Hence, there are
$$binom31binom21binom72binom52binom31binom21$$
such arrangements.



Both rows with one square are empty: There are $binom92$ ways to select two of the remaining nine positions for the $a$s, $binom72$ ways to choose two of the remaining seven positions for the $b$s, $binom51$ ways to choose one of the remaining five positions for the $c$, and $binom41$ ways to choose one the remaining four positions for the $d$. Hence, there are
$$binom92binom72binom51binom41$$
such arrangements.



Three empty rows: Since at most five squares can be left empty, the only way this can occur is if both of the rows with one square and one of the rows with three square are empty.



There are $binom31$ ways to select which of the rows with three squares is empty, $binom62$ ways to select two of the remaining six positions for the $a$s, $binom42$ ways to choose two of the remaining four positions for the $b$s, $binom21$ ways to choose one of the remaining two positions for the $c$, and $binom11$ ways to choose the remaining position for the $d$. Hence, there are
$$binom31binom62binom42binom21binom11$$
such arrangements.



It is not possible to have more than three empty rows. By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$binom112binom92binom71binom61 - binom31binom82binom62binom41binom31 - binom21binom102binom82binom61binom51 + binom31binom21binom72binom52binom31binom21 + binom92binom72binom51binom41 - binom31binom62binom42binom21binom11$$






share|cite|improve this answer















You have not accounted for the empty spaces in the grid.



There are $binom112$ ways to choose two of the eleven positions for the $a$s, $binom92$ ways to choose two of the remaining nine positions for the $b$s, $binom71$ ways to choose one of the remaining positions for the $c$, and $binom61$ ways to choose one of the remaining six positions for the $d$. Hence, you should have obtained
$$binom112binom92binom71binom61 = frac11!2!9! cdot frac9!2!7! cdot frac7!1!6! cdot frac6!1!5! = frac11!2!2!1!1!5!$$
distinguishable arrangements of two $a$s, two $b$s, one $c$, one $d$ in the grid without restriction. The factors of $2!$, $2!$, $1!$, $1!$, and $5!$ in the denominator account, respectively for the number of ways two $a$s, two $b$s, one $c$, one $d$, and five empty squares can be permuted among themselves without producing an arrangement that is distinguishable from the given arrangement. To see why, think of the problem as arranging two $a$s, two $b$s, one $c$, one $d$, and five $e$s in the eleven squares of the grid, where $e$ represents an empty square.



One empty row: The row can have either three squares or one square.



A row with three squares is empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom82$ ways to select two of the remaining eight positions for the $a$s, $binom62$ ways to choose two of the remaining six positions for the $b$s, $binom41$ ways to choose one of the remaining four positions for the $c$, and $binom31$ ways to choose one the remaining three positions for the $d$. Hence, there are
$$binom31binom82binom62binom41binom31$$
such arrangements.



A row with one square is empty: There are $binom21$ ways to choose which of the two rows with one square is empty, $binom102$ ways to select two of the remaining ten positions for the $a$s, $binom82$ ways to choose two of the remaining eight positions for the $b$s, $binom61$ ways to choose one of the remaining six positions for the $c$, and $binom51$ ways to choose one the remaining five positions for the $d$. Hence, there are
$$binom21binom102binom82binom61binom51$$
such arrangements.



Two empty rows: If two rows with three squares were empty, then there would not be a sufficient number of squares left in the grid for the six letters. Hence, we are left with two possibilities. Either a row with three squares and a row with one square are empty or both rows with one square are empty.



A row with three squares and a row with one square are empty: There are $binom31$ ways to choose which of the rows with three squares is empty, $binom21$ ways to choose which of the rows with two square are empty, $binom72$ ways to select two of the remaining seven positions for the $a$s, $binom52$ ways to choose two of the remaining five positions for the $b$s, $binom31$ ways to choose one of the remaining three positions for the $c$, and $binom21$ ways to choose one the remaining two positions for the $d$. Hence, there are
$$binom31binom21binom72binom52binom31binom21$$
such arrangements.



Both rows with one square are empty: There are $binom92$ ways to select two of the remaining nine positions for the $a$s, $binom72$ ways to choose two of the remaining seven positions for the $b$s, $binom51$ ways to choose one of the remaining five positions for the $c$, and $binom41$ ways to choose one the remaining four positions for the $d$. Hence, there are
$$binom92binom72binom51binom41$$
such arrangements.



Three empty rows: Since at most five squares can be left empty, the only way this can occur is if both of the rows with one square and one of the rows with three square are empty.



There are $binom31$ ways to select which of the rows with three squares is empty, $binom62$ ways to select two of the remaining six positions for the $a$s, $binom42$ ways to choose two of the remaining four positions for the $b$s, $binom21$ ways to choose one of the remaining two positions for the $c$, and $binom11$ ways to choose the remaining position for the $d$. Hence, there are
$$binom31binom62binom42binom21binom11$$
such arrangements.



It is not possible to have more than three empty rows. By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$binom112binom92binom71binom61 - binom31binom82binom62binom41binom31 - binom21binom102binom82binom61binom51 + binom31binom21binom72binom52binom31binom21 + binom92binom72binom51binom41 - binom31binom62binom42binom21binom11$$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Aug 3 at 10:54


























answered Aug 3 at 3:52









N. F. Taussig

37.9k93053




37.9k93053







  • 1




    Good one thanks a ton
    – Umesh shankar
    2 days ago












  • 1




    Good one thanks a ton
    – Umesh shankar
    2 days ago







1




1




Good one thanks a ton
– Umesh shankar
2 days ago




Good one thanks a ton
– Umesh shankar
2 days ago












 

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