The number of ways of choosing with parameters
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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 Feb 25 at 13:49
Vinyl_cape_jawa
3,34011433
asked Feb 25 at 4:56
cmplxlizcmplxliz
241
$endgroup$
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 Feb 25 at 13:49
Vinyl_cape_jawa
3,34011433
asked Feb 25 at 4:56
cmplxlizcmplxliz
241
$endgroup$
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 Feb 25 at 13:49
Vinyl_cape_jawa
3,34011433
asked Feb 25 at 4:56
cmplxlizcmplxliz
241
$endgroup$
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 Feb 25 at 13:49
Vinyl_cape_jawa
3,34011433
asked Feb 25 at 4:56
cmplxlizcmplxliz
241
edited Feb 25 at 13:49
Vinyl_cape_jawa
3,34011433
asked Feb 25 at 4:56
cmplxlizcmplxliz
241
edited Feb 25 at 13:49
Vinyl_cape_jawa
3,34011433
edited Feb 25 at 13:49
Vinyl_cape_jawa
3,34011433
edited Feb 25 at 13:49
Vinyl_cape_jawa
3,34011433
3,34011433
asked Feb 25 at 4:56
cmplxlizcmplxliz
241
asked Feb 25 at 4:56
cmplxlizcmplxliz
241
asked Feb 25 at 4:56
cmplxlizcmplxliz
241
241
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhboxandquad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered Feb 25 at 5:02
DavidDavid
69.5k668130
$endgroup$
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 Feb 25 at 5:28
Dbchatto67Dbchatto67
2,106320
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhboxandquad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered Feb 25 at 5:02
DavidDavid
69.5k668130
$endgroup$
add a comment |
$begingroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhboxandquad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered Feb 25 at 5:02
DavidDavid
69.5k668130
$endgroup$
add a comment |
$begingroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhboxandquad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered Feb 25 at 5:02
DavidDavid
69.5k668130
$endgroup$
Because under your scheme you would count, for example, both
$$R1,B1,Y1,R2quadhboxandquad R2,B1,Y1,R1 .$$
But these are actually the same choice and therefore should not be counted twice.
answered Feb 25 at 5:02
DavidDavid
69.5k668130
answered Feb 25 at 5:02
DavidDavid
69.5k668130
answered Feb 25 at 5:02
DavidDavid
69.5k668130
answered Feb 25 at 5:02
DavidDavid
69.5k668130
69.5k668130
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 Feb 25 at 5:28
Dbchatto67Dbchatto67
2,106320
$endgroup$
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 Feb 25 at 5:28
Dbchatto67Dbchatto67
2,106320
$endgroup$
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 Feb 25 at 5:28
Dbchatto67Dbchatto67
2,106320
$endgroup$
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 Feb 25 at 5:28
Dbchatto67Dbchatto67
2,106320
answered Feb 25 at 5:28
Dbchatto67Dbchatto67
2,106320
answered Feb 25 at 5:28
Dbchatto67Dbchatto67
2,106320
answered Feb 25 at 5:28
Dbchatto67Dbchatto67
2,106320
2,106320
add a comment |
add a comment |
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