Falling factorial counts permutations, what does rising factorial count?
Rising factorial example: Let $x = 7$ and $r = 4$. Then $7^{(4)} = 7(8)(9)(10) = 5040$. If we divide $7^{(4)}$ by $4!$ it counts multisubsets.
But what kind of combinatorial problem does rising factorial solve on its own?
discrete-mathematics
asked Jun 18 '15 at 1:11
Kyle
402
add a comment |
Rising factorial example: Let $x = 7$ and $r = 4$. Then $7^{(4)} = 7(8)(9)(10) = 5040$. If we divide $7^{(4)}$ by $4!$ it counts multisubsets.
But what kind of combinatorial problem does rising factorial solve on its own?
discrete-mathematics
asked Jun 18 '15 at 1:11
Kyle
402
Wouldn't rising subsets we used as the equivalent of Permutations?
– Kbot
Jun 18 '15 at 1:18
I am not sure. For example, $7! = 7 cdot 6 cdot ldots 1$.
– Kyle
Jun 18 '15 at 1:21
That's a factorial...
– Kbot
Jun 18 '15 at 2:43
add a comment |
Rising factorial example: Let $x = 7$ and $r = 4$. Then $7^{(4)} = 7(8)(9)(10) = 5040$. If we divide $7^{(4)}$ by $4!$ it counts multisubsets.
But what kind of combinatorial problem does rising factorial solve on its own?
discrete-mathematics
asked Jun 18 '15 at 1:11
Kyle
402
Rising factorial example: Let $x = 7$ and $r = 4$. Then $7^{(4)} = 7(8)(9)(10) = 5040$. If we divide $7^{(4)}$ by $4!$ it counts multisubsets.
But what kind of combinatorial problem does rising factorial solve on its own?
discrete-mathematics
discrete-mathematics
asked Jun 18 '15 at 1:11
Kyle
402
asked Jun 18 '15 at 1:11
Kyle
402
asked Jun 18 '15 at 1:11
Kyle
402
asked Jun 18 '15 at 1:11
Kyle
402
asked Jun 18 '15 at 1:11
Kyle
402
402
Wouldn't rising subsets we used as the equivalent of Permutations?
– Kbot
Jun 18 '15 at 1:18
I am not sure. For example, $7! = 7 cdot 6 cdot ldots 1$.
– Kyle
Jun 18 '15 at 1:21
That's a factorial...
– Kbot
Jun 18 '15 at 2:43
add a comment |
Wouldn't rising subsets we used as the equivalent of Permutations?
– Kbot
Jun 18 '15 at 1:18
I am not sure. For example, $7! = 7 cdot 6 cdot ldots 1$.
– Kyle
Jun 18 '15 at 1:21
That's a factorial...
– Kbot
Jun 18 '15 at 2:43
Wouldn't rising subsets we used as the equivalent of Permutations?
– Kbot
Jun 18 '15 at 1:18
Wouldn't rising subsets we used as the equivalent of Permutations?
– Kbot
Jun 18 '15 at 1:18
I am not sure. For example, $7! = 7 cdot 6 cdot ldots 1$.
– Kyle
Jun 18 '15 at 1:21
I am not sure. For example, $7! = 7 cdot 6 cdot ldots 1$.
– Kyle
Jun 18 '15 at 1:21
That's a factorial...
– Kbot
Jun 18 '15 at 2:43
That's a factorial...
– Kbot
Jun 18 '15 at 2:43
add a comment |
1 Answer
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Imagine constructing a permutation of $k$ elements with $n-1$ elements initially, but add a spot for each number placed in the sequence instead of taking one away. I'm going to call this a rising permutation for lack of a better name.
You can visualize the construction of an ordinary permutation on $n$ elements by imagining $n$ slots and replacing one slot with a number every turn. We'll number the turns $1,2, dots, n$ . In this array below each $*$ is an as-yet-unfilled box and each row represents the state at the end of the $n$th turn, with the first row being the initial state.
$$
begin{array}{ccc}
* & * & * \
* & 1 & * \
2 & 1 & * \
2 & 1 & 3 \
end{array}
$$
There are several equivalent ways of thinking about constructing a permutation one element at a time. This way doesn't make it easy to visualize what happens if there are more numbers than slots (the falling factorial case). However, it does make the analogy with a "rising permutation" more obvious. The key difference is that, on the $i$th turn, we place the number $i$ between one of the items already in the sequence. Doing so creates a slot instead of consuming one.
Each row represents a turn. The gaps in the earlier rows are so that each column tracks the lifetime of a single item in the sequence.
$$ begin{array}{cccccc}
& * & * & & * & \
& * & * & 1 & * & \
2 & * & * & 1 & * & \
2 & * & * & 1 & * & 3 \
end{array}$$
The above example is from $(4)^{overline{3}}$ .
The rising factorial $(n)^{overline{k}}$ asks how many distinct sequences can be made after $k$ turns if there are $(n-1)$ stars initially (equivalently $n$ stars at the end of turn $1$) . Note that a sequence contains the stars that are in it so * 1 * and 1 * * are not the same sequence.
answered Nov 19 '18 at 2:05
Gregory Nisbet
524312
add a comment |
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1 Answer
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1 Answer
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Imagine constructing a permutation of $k$ elements with $n-1$ elements initially, but add a spot for each number placed in the sequence instead of taking one away. I'm going to call this a rising permutation for lack of a better name.
You can visualize the construction of an ordinary permutation on $n$ elements by imagining $n$ slots and replacing one slot with a number every turn. We'll number the turns $1,2, dots, n$ . In this array below each $*$ is an as-yet-unfilled box and each row represents the state at the end of the $n$th turn, with the first row being the initial state.
$$
begin{array}{ccc}
* & * & * \
* & 1 & * \
2 & 1 & * \
2 & 1 & 3 \
end{array}
$$
There are several equivalent ways of thinking about constructing a permutation one element at a time. This way doesn't make it easy to visualize what happens if there are more numbers than slots (the falling factorial case). However, it does make the analogy with a "rising permutation" more obvious. The key difference is that, on the $i$th turn, we place the number $i$ between one of the items already in the sequence. Doing so creates a slot instead of consuming one.
Each row represents a turn. The gaps in the earlier rows are so that each column tracks the lifetime of a single item in the sequence.
$$ begin{array}{cccccc}
& * & * & & * & \
& * & * & 1 & * & \
2 & * & * & 1 & * & \
2 & * & * & 1 & * & 3 \
end{array}$$
The above example is from $(4)^{overline{3}}$ .
The rising factorial $(n)^{overline{k}}$ asks how many distinct sequences can be made after $k$ turns if there are $(n-1)$ stars initially (equivalently $n$ stars at the end of turn $1$) . Note that a sequence contains the stars that are in it so * 1 * and 1 * * are not the same sequence.
answered Nov 19 '18 at 2:05
Gregory Nisbet
524312
add a comment |
Imagine constructing a permutation of $k$ elements with $n-1$ elements initially, but add a spot for each number placed in the sequence instead of taking one away. I'm going to call this a rising permutation for lack of a better name.
You can visualize the construction of an ordinary permutation on $n$ elements by imagining $n$ slots and replacing one slot with a number every turn. We'll number the turns $1,2, dots, n$ . In this array below each $*$ is an as-yet-unfilled box and each row represents the state at the end of the $n$th turn, with the first row being the initial state.
$$
begin{array}{ccc}
* & * & * \
* & 1 & * \
2 & 1 & * \
2 & 1 & 3 \
end{array}
$$
There are several equivalent ways of thinking about constructing a permutation one element at a time. This way doesn't make it easy to visualize what happens if there are more numbers than slots (the falling factorial case). However, it does make the analogy with a "rising permutation" more obvious. The key difference is that, on the $i$th turn, we place the number $i$ between one of the items already in the sequence. Doing so creates a slot instead of consuming one.
Each row represents a turn. The gaps in the earlier rows are so that each column tracks the lifetime of a single item in the sequence.
$$ begin{array}{cccccc}
& * & * & & * & \
& * & * & 1 & * & \
2 & * & * & 1 & * & \
2 & * & * & 1 & * & 3 \
end{array}$$
The above example is from $(4)^{overline{3}}$ .
The rising factorial $(n)^{overline{k}}$ asks how many distinct sequences can be made after $k$ turns if there are $(n-1)$ stars initially (equivalently $n$ stars at the end of turn $1$) . Note that a sequence contains the stars that are in it so * 1 * and 1 * * are not the same sequence.
answered Nov 19 '18 at 2:05
Gregory Nisbet
524312
add a comment |
Imagine constructing a permutation of $k$ elements with $n-1$ elements initially, but add a spot for each number placed in the sequence instead of taking one away. I'm going to call this a rising permutation for lack of a better name.
You can visualize the construction of an ordinary permutation on $n$ elements by imagining $n$ slots and replacing one slot with a number every turn. We'll number the turns $1,2, dots, n$ . In this array below each $*$ is an as-yet-unfilled box and each row represents the state at the end of the $n$th turn, with the first row being the initial state.
$$
begin{array}{ccc}
* & * & * \
* & 1 & * \
2 & 1 & * \
2 & 1 & 3 \
end{array}
$$
There are several equivalent ways of thinking about constructing a permutation one element at a time. This way doesn't make it easy to visualize what happens if there are more numbers than slots (the falling factorial case). However, it does make the analogy with a "rising permutation" more obvious. The key difference is that, on the $i$th turn, we place the number $i$ between one of the items already in the sequence. Doing so creates a slot instead of consuming one.
Each row represents a turn. The gaps in the earlier rows are so that each column tracks the lifetime of a single item in the sequence.
$$ begin{array}{cccccc}
& * & * & & * & \
& * & * & 1 & * & \
2 & * & * & 1 & * & \
2 & * & * & 1 & * & 3 \
end{array}$$
The above example is from $(4)^{overline{3}}$ .
The rising factorial $(n)^{overline{k}}$ asks how many distinct sequences can be made after $k$ turns if there are $(n-1)$ stars initially (equivalently $n$ stars at the end of turn $1$) . Note that a sequence contains the stars that are in it so * 1 * and 1 * * are not the same sequence.
answered Nov 19 '18 at 2:05
Gregory Nisbet
524312
Imagine constructing a permutation of $k$ elements with $n-1$ elements initially, but add a spot for each number placed in the sequence instead of taking one away. I'm going to call this a rising permutation for lack of a better name.
You can visualize the construction of an ordinary permutation on $n$ elements by imagining $n$ slots and replacing one slot with a number every turn. We'll number the turns $1,2, dots, n$ . In this array below each $*$ is an as-yet-unfilled box and each row represents the state at the end of the $n$th turn, with the first row being the initial state.
$$
begin{array}{ccc}
* & * & * \
* & 1 & * \
2 & 1 & * \
2 & 1 & 3 \
end{array}
$$
There are several equivalent ways of thinking about constructing a permutation one element at a time. This way doesn't make it easy to visualize what happens if there are more numbers than slots (the falling factorial case). However, it does make the analogy with a "rising permutation" more obvious. The key difference is that, on the $i$th turn, we place the number $i$ between one of the items already in the sequence. Doing so creates a slot instead of consuming one.
Each row represents a turn. The gaps in the earlier rows are so that each column tracks the lifetime of a single item in the sequence.
$$ begin{array}{cccccc}
& * & * & & * & \
& * & * & 1 & * & \
2 & * & * & 1 & * & \
2 & * & * & 1 & * & 3 \
end{array}$$
The above example is from $(4)^{overline{3}}$ .
The rising factorial $(n)^{overline{k}}$ asks how many distinct sequences can be made after $k$ turns if there are $(n-1)$ stars initially (equivalently $n$ stars at the end of turn $1$) . Note that a sequence contains the stars that are in it so * 1 * and 1 * * are not the same sequence.
answered Nov 19 '18 at 2:05
Gregory Nisbet
524312
answered Nov 19 '18 at 2:05
Gregory Nisbet
524312
answered Nov 19 '18 at 2:05
Gregory Nisbet
524312
answered Nov 19 '18 at 2:05
Gregory Nisbet
524312
524312
add a comment |
add a comment |
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Wouldn't rising subsets we used as the equivalent of Permutations?
– Kbot
Jun 18 '15 at 1:18
I am not sure. For example, $7! = 7 cdot 6 cdot ldots 1$.
– Kyle
Jun 18 '15 at 1:21
That's a factorial...
– Kbot
Jun 18 '15 at 2:43