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The number of ways of choosing with parameters
Different ways of coloring a $4 times 4$ game boardWhat does the “n choose multiple numbers” symbol stands for?Recurrence Relation, Discrete Math problem(Homework)Condition over CombinationsWith how many different ways can Adriana be dressed…????Counting the number of trials.In how many ways we can color $15$ eggs..In how many ways can two different colored balls be chosen?Combinatorics: Coloring a prismProbability of picking two marbles each from two colors when selecting $4$ marbles out of $30$ marbles
$begingroup$
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 $9cdot 10cdot 11cdot 30$.
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
edited 1 hour ago
Vinyl_coat_jawa
3,0101132
asked 10 hours ago
cmplxlizcmplxliz
211
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cmplxliz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
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 $9cdot 10cdot 11cdot 30$.
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
edited 1 hour ago
Vinyl_coat_jawa
3,0101132
asked 10 hours ago
cmplxlizcmplxliz
211
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 |
$begingroup$
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 $9cdot 10cdot 11cdot 30$.
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
edited 1 hour ago
Vinyl_coat_jawa
3,0101132
asked 10 hours ago
cmplxlizcmplxliz
211
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.
$endgroup$
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 $9cdot 10cdot 11cdot 30$.
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
edited 1 hour ago
Vinyl_coat_jawa
3,0101132
asked 10 hours ago
cmplxlizcmplxliz
211
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.
edited 1 hour ago
Vinyl_coat_jawa
3,0101132
asked 10 hours ago
cmplxlizcmplxliz
211
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.
edited 1 hour ago
Vinyl_coat_jawa
3,0101132
edited 1 hour ago
Vinyl_coat_jawa
3,0101132
edited 1 hour ago
Vinyl_coat_jawa
3,0101132
3,0101132
asked 10 hours ago
cmplxlizcmplxliz
211
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 10 hours ago
cmplxlizcmplxliz
211
asked 10 hours ago
cmplxlizcmplxliz
211
211
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.
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add a comment |
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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 10 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 10 hours ago
Dbchatto67Dbchatto67
1,142118
$endgroup$
add a comment |
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2 Answers
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oldest
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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 10 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 10 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 10 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 10 hours ago
DavidDavid
69.2k667130
answered 10 hours ago
DavidDavid
69.2k667130
answered 10 hours ago
DavidDavid
69.2k667130
answered 10 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 10 hours ago
Dbchatto67Dbchatto67
1,142118
$endgroup$
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 10 hours ago
Dbchatto67Dbchatto67
1,142118
$endgroup$
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 10 hours ago
Dbchatto67Dbchatto67
1,142118
$endgroup$
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 10 hours ago
Dbchatto67Dbchatto67
1,142118
answered 10 hours ago
Dbchatto67Dbchatto67
1,142118
answered 10 hours ago
Dbchatto67Dbchatto67
1,142118
answered 10 hours ago
Dbchatto67Dbchatto67
1,142118
1,142118
add a comment |
add a comment |
cmplxliz is a new contributor. Be nice, and check out our Code of Conduct.
cmplxliz is a new contributor. Be nice, and check out our Code of Conduct.
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