Number of ways to choose tuples
I have six numbers: 1,2,3,4,5,6
I need to count in how many ways (I expect every way will have 5 tuples). Difference of two tuple members must not be bigger than 1.
For example:
(1,2) (2,3) (3,4) (4,5) (5,6) // this is one way
Is there any way to count this?
I need algorithm to write all possible ways.
Many tnx!
combinatorics
asked Nov 24 at 14:38
EnthusiasticPerson
31
add a comment |
I have six numbers: 1,2,3,4,5,6
I need to count in how many ways (I expect every way will have 5 tuples). Difference of two tuple members must not be bigger than 1.
For example:
(1,2) (2,3) (3,4) (4,5) (5,6) // this is one way
Is there any way to count this?
I need algorithm to write all possible ways.
Many tnx!
combinatorics
asked Nov 24 at 14:38
EnthusiasticPerson
31
add a comment |
I have six numbers: 1,2,3,4,5,6
I need to count in how many ways (I expect every way will have 5 tuples). Difference of two tuple members must not be bigger than 1.
For example:
(1,2) (2,3) (3,4) (4,5) (5,6) // this is one way
Is there any way to count this?
I need algorithm to write all possible ways.
Many tnx!
combinatorics
asked Nov 24 at 14:38
EnthusiasticPerson
31
I have six numbers: 1,2,3,4,5,6
I need to count in how many ways (I expect every way will have 5 tuples). Difference of two tuple members must not be bigger than 1.
For example:
(1,2) (2,3) (3,4) (4,5) (5,6) // this is one way
Is there any way to count this?
I need algorithm to write all possible ways.
Many tnx!
combinatorics
combinatorics
asked Nov 24 at 14:38
EnthusiasticPerson
31
asked Nov 24 at 14:38
EnthusiasticPerson
31
asked Nov 24 at 14:38
EnthusiasticPerson
31
asked Nov 24 at 14:38
EnthusiasticPerson
31
asked Nov 24 at 14:38
EnthusiasticPerson
31
31
add a comment |
add a comment |
1 Answer
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1) Unordered tuples of the form $(a, a+1)$, where $a$ and $a+1$ belong to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n-1$ ways; $a$ has to come from ${m_1, m_1+1, m_1+2, ..., m_1+n-2}$. Therefore, you get $n-1$ unordered tuples.
2) Unordered tuples of the form $(a, a)$, where $a$ belongs to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n$ ways. Therefore, you get $n$ tuples.
The total number of unordered tuples is $n+n-1=2n-1$. Note that if the tuples are ordered, we can reverse each tuple of case (1) to get another admissible tuple. This process doesn't generate new tuples for any tuple of case (2). The total tuples will then be $n+2(n-1)=3n-2$.
answered Nov 24 at 15:05
Shubham Johri
3,826716
add a comment |
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1 Answer
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1 Answer
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1) Unordered tuples of the form $(a, a+1)$, where $a$ and $a+1$ belong to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n-1$ ways; $a$ has to come from ${m_1, m_1+1, m_1+2, ..., m_1+n-2}$. Therefore, you get $n-1$ unordered tuples.
2) Unordered tuples of the form $(a, a)$, where $a$ belongs to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n$ ways. Therefore, you get $n$ tuples.
The total number of unordered tuples is $n+n-1=2n-1$. Note that if the tuples are ordered, we can reverse each tuple of case (1) to get another admissible tuple. This process doesn't generate new tuples for any tuple of case (2). The total tuples will then be $n+2(n-1)=3n-2$.
answered Nov 24 at 15:05
Shubham Johri
3,826716
add a comment |
1) Unordered tuples of the form $(a, a+1)$, where $a$ and $a+1$ belong to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n-1$ ways; $a$ has to come from ${m_1, m_1+1, m_1+2, ..., m_1+n-2}$. Therefore, you get $n-1$ unordered tuples.
2) Unordered tuples of the form $(a, a)$, where $a$ belongs to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n$ ways. Therefore, you get $n$ tuples.
The total number of unordered tuples is $n+n-1=2n-1$. Note that if the tuples are ordered, we can reverse each tuple of case (1) to get another admissible tuple. This process doesn't generate new tuples for any tuple of case (2). The total tuples will then be $n+2(n-1)=3n-2$.
answered Nov 24 at 15:05
Shubham Johri
3,826716
add a comment |
1) Unordered tuples of the form $(a, a+1)$, where $a$ and $a+1$ belong to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n-1$ ways; $a$ has to come from ${m_1, m_1+1, m_1+2, ..., m_1+n-2}$. Therefore, you get $n-1$ unordered tuples.
2) Unordered tuples of the form $(a, a)$, where $a$ belongs to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n$ ways. Therefore, you get $n$ tuples.
The total number of unordered tuples is $n+n-1=2n-1$. Note that if the tuples are ordered, we can reverse each tuple of case (1) to get another admissible tuple. This process doesn't generate new tuples for any tuple of case (2). The total tuples will then be $n+2(n-1)=3n-2$.
answered Nov 24 at 15:05
Shubham Johri
3,826716
1) Unordered tuples of the form $(a, a+1)$, where $a$ and $a+1$ belong to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n-1$ ways; $a$ has to come from ${m_1, m_1+1, m_1+2, ..., m_1+n-2}$. Therefore, you get $n-1$ unordered tuples.
2) Unordered tuples of the form $(a, a)$, where $a$ belongs to the list of $n$ consecutive integers, ${m_1, m_1+1, m_1+2, ..., m_1+n-1}$: You can select $a$ in $n$ ways. Therefore, you get $n$ tuples.
The total number of unordered tuples is $n+n-1=2n-1$. Note that if the tuples are ordered, we can reverse each tuple of case (1) to get another admissible tuple. This process doesn't generate new tuples for any tuple of case (2). The total tuples will then be $n+2(n-1)=3n-2$.
answered Nov 24 at 15:05
Shubham Johri
3,826716
edited Nov 24 at 15:16
answered Nov 24 at 15:05
Shubham Johri
3,826716
answered Nov 24 at 15:05
Shubham Johri
3,826716
answered Nov 24 at 15:05
Shubham Johri
3,826716
3,826716
add a comment |
add a comment |
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