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Coombinatorics- The number of ways of choosing with parameters
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Coombinatorics- The number of ways of choosing with parameters
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$begingroup$
Here's the problem:
At our disposal is a collection of 10 red, 11 blue and 12 yellow fabrics. (each fabric is unique)In how many ways can we choose 4 different fabrics if we want at least one fabric of each of the three colors?
My solution was since the first fabric chosen must be red, there are 10 options for it. Then the next fabric must be blue, which has 11 options. The third fabric is yellow, with 12 options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are (9+10+11-3)= 30 ways to choose the last one, making the total number of choices 9*10*11*30.
My professor said that I needed to divide that by 2 to get the right answer, but I just don't understand why. Any help would be much appreciated!
combinatorics discrete-mathematics
asked 4 hours ago
cmplxlizcmplxliz
111
New contributor
cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Here's the problem:
At our disposal is a collection of 10 red, 11 blue and 12 yellow fabrics. (each fabric is unique)In how many ways can we choose 4 different fabrics if we want at least one fabric of each of the three colors?
My solution was since the first fabric chosen must be red, there are 10 options for it. Then the next fabric must be blue, which has 11 options. The third fabric is yellow, with 12 options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are (9+10+11-3)= 30 ways to choose the last one, making the total number of choices 9*10*11*30.
My professor said that I needed to divide that by 2 to get the right answer, but I just don't understand why. Any help would be much appreciated!
combinatorics discrete-mathematics
asked 4 hours ago
cmplxlizcmplxliz
111
New contributor
cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Here's the problem:
At our disposal is a collection of 10 red, 11 blue and 12 yellow fabrics. (each fabric is unique)In how many ways can we choose 4 different fabrics if we want at least one fabric of each of the three colors?
My solution was since the first fabric chosen must be red, there are 10 options for it. Then the next fabric must be blue, which has 11 options. The third fabric is yellow, with 12 options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are (9+10+11-3)= 30 ways to choose the last one, making the total number of choices 9*10*11*30.
My professor said that I needed to divide that by 2 to get the right answer, but I just don't understand why. Any help would be much appreciated!
combinatorics discrete-mathematics
asked 4 hours ago
cmplxlizcmplxliz
111
New contributor
cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Here's the problem:
At our disposal is a collection of 10 red, 11 blue and 12 yellow fabrics. (each fabric is unique)In how many ways can we choose 4 different fabrics if we want at least one fabric of each of the three colors?
My solution was since the first fabric chosen must be red, there are 10 options for it. Then the next fabric must be blue, which has 11 options. The third fabric is yellow, with 12 options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are (9+10+11-3)= 30 ways to choose the last one, making the total number of choices 9*10*11*30.
My professor said that I needed to divide that by 2 to get the right answer, but I just don't understand why. Any help would be much appreciated!
combinatorics discrete-mathematics
combinatorics discrete-mathematics
asked 4 hours ago
cmplxlizcmplxliz
111
New contributor
cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
cmplxlizcmplxliz
111
New contributor
cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
cmplxlizcmplxliz
111
New contributor
cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
cmplxlizcmplxliz
111
asked 4 hours ago
cmplxlizcmplxliz
111
111
New contributor
cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
cmplxliz 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
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$begingroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered 4 hours ago
DavidDavid
69.2k667130
$endgroup$
add a comment |
$begingroup$
The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$
answered 3 hours ago
Dbchatto67Dbchatto67
1,134118
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered 4 hours ago
DavidDavid
69.2k667130
$endgroup$
add a comment |
$begingroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered 4 hours ago
DavidDavid
69.2k667130
$endgroup$
add a comment |
$begingroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered 4 hours ago
DavidDavid
69.2k667130
$endgroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhbox{and}quad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered 4 hours ago
DavidDavid
69.2k667130
answered 4 hours ago
DavidDavid
69.2k667130
answered 4 hours ago
DavidDavid
69.2k667130
answered 4 hours ago
DavidDavid
69.2k667130
69.2k667130
add a comment |
add a comment |
$begingroup$
The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$
answered 3 hours ago
Dbchatto67Dbchatto67
1,134118
$endgroup$
add a comment |
$begingroup$
The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$
answered 3 hours ago
Dbchatto67Dbchatto67
1,134118
$endgroup$
add a comment |
$begingroup$
The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$
answered 3 hours ago
Dbchatto67Dbchatto67
1,134118
$endgroup$
The total number of ways is $binom {10} {2} binom {11} {1} binom {12} {1} + binom {10} {1} binom {11} {2} binom {12} {1} + binom {10} {1} binom {11} {1} binom {12} {2} = 19800.$
answered 3 hours ago
Dbchatto67Dbchatto67
1,134118
answered 3 hours ago
Dbchatto67Dbchatto67
1,134118
answered 3 hours ago
Dbchatto67Dbchatto67
1,134118
answered 3 hours ago
Dbchatto67Dbchatto67
1,134118
1,134118
add a comment |
add a comment |
cmplxliz is a new contributor. Be nice, and check out our Code of Conduct.
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cmplxliz is a new contributor. Be nice, and check out our Code of Conduct.
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