Combinatorics - selecting from distinct sets
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We have $6$ items of class $A$, $4$ of class $B$, and $2$ of class $C$. How many distinct three-item sets can we select?
It's easy to see that the answer is $9$ by listing the possibilities:
${AAA}$, ${BBB}$, ${AAB}$, ${AAC}$, ${BBA}$, ${BBC}$, ${CCA}$, ${CCB}$, ${ABC}$
However, I would like to determine the answer in a more principled manner and I am having trouble doing so.
combinatorics discrete-mathematics
edited Nov 17 at 7:12
ArsenBerk
7,69131338
asked Nov 17 at 6:57
Gaussian0617
584318
add a comment |
up vote
1
down vote
favorite
We have $6$ items of class $A$, $4$ of class $B$, and $2$ of class $C$. How many distinct three-item sets can we select?
It's easy to see that the answer is $9$ by listing the possibilities:
${AAA}$, ${BBB}$, ${AAB}$, ${AAC}$, ${BBA}$, ${BBC}$, ${CCA}$, ${CCB}$, ${ABC}$
However, I would like to determine the answer in a more principled manner and I am having trouble doing so.
combinatorics discrete-mathematics
edited Nov 17 at 7:12
ArsenBerk
7,69131338
asked Nov 17 at 6:57
Gaussian0617
584318
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We have $6$ items of class $A$, $4$ of class $B$, and $2$ of class $C$. How many distinct three-item sets can we select?
It's easy to see that the answer is $9$ by listing the possibilities:
${AAA}$, ${BBB}$, ${AAB}$, ${AAC}$, ${BBA}$, ${BBC}$, ${CCA}$, ${CCB}$, ${ABC}$
However, I would like to determine the answer in a more principled manner and I am having trouble doing so.
combinatorics discrete-mathematics
edited Nov 17 at 7:12
ArsenBerk
7,69131338
asked Nov 17 at 6:57
Gaussian0617
584318
We have $6$ items of class $A$, $4$ of class $B$, and $2$ of class $C$. How many distinct three-item sets can we select?
It's easy to see that the answer is $9$ by listing the possibilities:
${AAA}$, ${BBB}$, ${AAB}$, ${AAC}$, ${BBA}$, ${BBC}$, ${CCA}$, ${CCB}$, ${ABC}$
However, I would like to determine the answer in a more principled manner and I am having trouble doing so.
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Nov 17 at 7:12
ArsenBerk
7,69131338
asked Nov 17 at 6:57
Gaussian0617
584318
edited Nov 17 at 7:12
ArsenBerk
7,69131338
asked Nov 17 at 6:57
Gaussian0617
584318
edited Nov 17 at 7:12
ArsenBerk
7,69131338
edited Nov 17 at 7:12
ArsenBerk
7,69131338
edited Nov 17 at 7:12
ArsenBerk
7,69131338
7,69131338
asked Nov 17 at 6:57
Gaussian0617
584318
asked Nov 17 at 6:57
Gaussian0617
584318
asked Nov 17 at 6:57
Gaussian0617
584318
584318
add a comment |
add a comment |
1 Answer
1
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1
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If we have three distinct items, we should choose three classes by $binom{3}{3}$ so we have $1$ possibility here.
If we have two distinct items, we should choose two classes by $binom{3}{2}$ and we can also interchange the items by $2$ ways (like two from $A$, one from $B$ and two from $B$, one from $A$) so we have $2binom{3}{2} = 6$ possibilities here.
If we have only one type of item, we should choose either the first class or second class (since class $C$ doesn not have three items in it) so we have $2$ possiblities here.
In total, we have $1+6+2 = 9$ possibilities.
answered Nov 17 at 7:11
ArsenBerk
7,69131338
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If we have three distinct items, we should choose three classes by $binom{3}{3}$ so we have $1$ possibility here.
If we have two distinct items, we should choose two classes by $binom{3}{2}$ and we can also interchange the items by $2$ ways (like two from $A$, one from $B$ and two from $B$, one from $A$) so we have $2binom{3}{2} = 6$ possibilities here.
If we have only one type of item, we should choose either the first class or second class (since class $C$ doesn not have three items in it) so we have $2$ possiblities here.
In total, we have $1+6+2 = 9$ possibilities.
answered Nov 17 at 7:11
ArsenBerk
7,69131338
add a comment |
up vote
1
down vote
accepted
If we have three distinct items, we should choose three classes by $binom{3}{3}$ so we have $1$ possibility here.
If we have two distinct items, we should choose two classes by $binom{3}{2}$ and we can also interchange the items by $2$ ways (like two from $A$, one from $B$ and two from $B$, one from $A$) so we have $2binom{3}{2} = 6$ possibilities here.
If we have only one type of item, we should choose either the first class or second class (since class $C$ doesn not have three items in it) so we have $2$ possiblities here.
In total, we have $1+6+2 = 9$ possibilities.
answered Nov 17 at 7:11
ArsenBerk
7,69131338
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If we have three distinct items, we should choose three classes by $binom{3}{3}$ so we have $1$ possibility here.
If we have two distinct items, we should choose two classes by $binom{3}{2}$ and we can also interchange the items by $2$ ways (like two from $A$, one from $B$ and two from $B$, one from $A$) so we have $2binom{3}{2} = 6$ possibilities here.
If we have only one type of item, we should choose either the first class or second class (since class $C$ doesn not have three items in it) so we have $2$ possiblities here.
In total, we have $1+6+2 = 9$ possibilities.
answered Nov 17 at 7:11
ArsenBerk
7,69131338
If we have three distinct items, we should choose three classes by $binom{3}{3}$ so we have $1$ possibility here.
If we have two distinct items, we should choose two classes by $binom{3}{2}$ and we can also interchange the items by $2$ ways (like two from $A$, one from $B$ and two from $B$, one from $A$) so we have $2binom{3}{2} = 6$ possibilities here.
If we have only one type of item, we should choose either the first class or second class (since class $C$ doesn not have three items in it) so we have $2$ possiblities here.
In total, we have $1+6+2 = 9$ possibilities.
answered Nov 17 at 7:11
ArsenBerk
7,69131338
answered Nov 17 at 7:11
ArsenBerk
7,69131338
answered Nov 17 at 7:11
ArsenBerk
7,69131338
answered Nov 17 at 7:11
ArsenBerk
7,69131338
7,69131338
add a comment |
add a comment |
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