Mixing
Enumerating 5 distinct elements into 3 subsets
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What is manual way of enumerations of 5 items into 3 different subsets.
Enumeration of any number of items mechanically into subsets of size 1 is just each element on it's own.
Enumerating into subsets of size 2 is just picking one of the elements from original set and then picking another elements.
enumerating into subsets of size n, where n is less than or equal to size of original set can be done recursively.
But cant seem to find a similar mechanical way to enumerate all the possible 3 subsets of 5 elements.
combinatorics
asked Jan 29 at 7:46
ArjangArjang
5,65462364
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add a comment |
$begingroup$
What is manual way of enumerations of 5 items into 3 different subsets.
Enumeration of any number of items mechanically into subsets of size 1 is just each element on it's own.
Enumerating into subsets of size 2 is just picking one of the elements from original set and then picking another elements.
enumerating into subsets of size n, where n is less than or equal to size of original set can be done recursively.
But cant seem to find a similar mechanical way to enumerate all the possible 3 subsets of 5 elements.
combinatorics
asked Jan 29 at 7:46
ArjangArjang
5,65462364
$endgroup$
add a comment |
$begingroup$
What is manual way of enumerations of 5 items into 3 different subsets.
Enumeration of any number of items mechanically into subsets of size 1 is just each element on it's own.
Enumerating into subsets of size 2 is just picking one of the elements from original set and then picking another elements.
enumerating into subsets of size n, where n is less than or equal to size of original set can be done recursively.
But cant seem to find a similar mechanical way to enumerate all the possible 3 subsets of 5 elements.
combinatorics
asked Jan 29 at 7:46
ArjangArjang
5,65462364
$endgroup$
What is manual way of enumerations of 5 items into 3 different subsets.
Enumeration of any number of items mechanically into subsets of size 1 is just each element on it's own.
Enumerating into subsets of size 2 is just picking one of the elements from original set and then picking another elements.
enumerating into subsets of size n, where n is less than or equal to size of original set can be done recursively.
But cant seem to find a similar mechanical way to enumerate all the possible 3 subsets of 5 elements.
combinatorics
combinatorics
asked Jan 29 at 7:46
ArjangArjang
5,65462364
asked Jan 29 at 7:46
ArjangArjang
5,65462364
asked Jan 29 at 7:46
ArjangArjang
5,65462364
asked Jan 29 at 7:46
ArjangArjang
5,65462364
asked Jan 29 at 7:46
ArjangArjang
5,65462364
5,65462364
add a comment |
add a comment |
1 Answer
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To divide five items into three non-empty subsets, you could divide them $3-1-1$ or $2-2-1$. There are ${5 choose 3} = 10$ possible ways to handle the first case, and ${5 choose 4}{3 choose 1} = 15$ to handle the second case. Considering the numbers 1 through 5, the first case would contain:
$${1, 2, 3}, {4}, {5}$$
$${1, 2, 4}, {3}, {5}$$
$${1, 2, 5}, {3}, {4}$$
$${1, 3, 4}, {2}, {5}$$
$$ldots$$
$${3, 4, 5}, {1}, {2}$$
The second case would then contain:
$${1, 2}, {3, 4}, {5}$$
$${1, 2}, {3, 5}, {4}$$
$${1, 2}, {4, 5}, {3}$$
$${1, 3}, {2, 4}, {5}$$
$$ldots$$
$${2, 5}, {3, 4}, {1}$$
answered Jan 29 at 7:50
jvdhooftjvdhooft
5,67561641
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1 Answer
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1 Answer
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$begingroup$
To divide five items into three non-empty subsets, you could divide them $3-1-1$ or $2-2-1$. There are ${5 choose 3} = 10$ possible ways to handle the first case, and ${5 choose 4}{3 choose 1} = 15$ to handle the second case. Considering the numbers 1 through 5, the first case would contain:
$${1, 2, 3}, {4}, {5}$$
$${1, 2, 4}, {3}, {5}$$
$${1, 2, 5}, {3}, {4}$$
$${1, 3, 4}, {2}, {5}$$
$$ldots$$
$${3, 4, 5}, {1}, {2}$$
The second case would then contain:
$${1, 2}, {3, 4}, {5}$$
$${1, 2}, {3, 5}, {4}$$
$${1, 2}, {4, 5}, {3}$$
$${1, 3}, {2, 4}, {5}$$
$$ldots$$
$${2, 5}, {3, 4}, {1}$$
answered Jan 29 at 7:50
jvdhooftjvdhooft
5,67561641
$endgroup$
add a comment |
$begingroup$
To divide five items into three non-empty subsets, you could divide them $3-1-1$ or $2-2-1$. There are ${5 choose 3} = 10$ possible ways to handle the first case, and ${5 choose 4}{3 choose 1} = 15$ to handle the second case. Considering the numbers 1 through 5, the first case would contain:
$${1, 2, 3}, {4}, {5}$$
$${1, 2, 4}, {3}, {5}$$
$${1, 2, 5}, {3}, {4}$$
$${1, 3, 4}, {2}, {5}$$
$$ldots$$
$${3, 4, 5}, {1}, {2}$$
The second case would then contain:
$${1, 2}, {3, 4}, {5}$$
$${1, 2}, {3, 5}, {4}$$
$${1, 2}, {4, 5}, {3}$$
$${1, 3}, {2, 4}, {5}$$
$$ldots$$
$${2, 5}, {3, 4}, {1}$$
answered Jan 29 at 7:50
jvdhooftjvdhooft
5,67561641
$endgroup$
add a comment |
$begingroup$
To divide five items into three non-empty subsets, you could divide them $3-1-1$ or $2-2-1$. There are ${5 choose 3} = 10$ possible ways to handle the first case, and ${5 choose 4}{3 choose 1} = 15$ to handle the second case. Considering the numbers 1 through 5, the first case would contain:
$${1, 2, 3}, {4}, {5}$$
$${1, 2, 4}, {3}, {5}$$
$${1, 2, 5}, {3}, {4}$$
$${1, 3, 4}, {2}, {5}$$
$$ldots$$
$${3, 4, 5}, {1}, {2}$$
The second case would then contain:
$${1, 2}, {3, 4}, {5}$$
$${1, 2}, {3, 5}, {4}$$
$${1, 2}, {4, 5}, {3}$$
$${1, 3}, {2, 4}, {5}$$
$$ldots$$
$${2, 5}, {3, 4}, {1}$$
answered Jan 29 at 7:50
jvdhooftjvdhooft
5,67561641
$endgroup$
To divide five items into three non-empty subsets, you could divide them $3-1-1$ or $2-2-1$. There are ${5 choose 3} = 10$ possible ways to handle the first case, and ${5 choose 4}{3 choose 1} = 15$ to handle the second case. Considering the numbers 1 through 5, the first case would contain:
$${1, 2, 3}, {4}, {5}$$
$${1, 2, 4}, {3}, {5}$$
$${1, 2, 5}, {3}, {4}$$
$${1, 3, 4}, {2}, {5}$$
$$ldots$$
$${3, 4, 5}, {1}, {2}$$
The second case would then contain:
$${1, 2}, {3, 4}, {5}$$
$${1, 2}, {3, 5}, {4}$$
$${1, 2}, {4, 5}, {3}$$
$${1, 3}, {2, 4}, {5}$$
$$ldots$$
$${2, 5}, {3, 4}, {1}$$
answered Jan 29 at 7:50
jvdhooftjvdhooft
5,67561641
answered Jan 29 at 7:50
jvdhooftjvdhooft
5,67561641
answered Jan 29 at 7:50
jvdhooftjvdhooft
5,67561641
answered Jan 29 at 7:50
jvdhooftjvdhooft
5,67561641
5,67561641
add a comment |
add a comment |
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