Falling factorial counts permutations, what does rising factorial count?












2














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







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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
















2














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







share|cite|improve this question






















  • 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














2












2








2







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










1 Answer
1






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oldest

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0














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.






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    1 Answer
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    1 Answer
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    0














    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.






    share|cite|improve this answer


























      0














      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.






      share|cite|improve this answer
























        0












        0








        0






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 '18 at 2:05









        Gregory Nisbet

        524312




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