Integers $1,2,…,n$ are placed in a way that each value is either bigger or smaller than all preceding...












0












$begingroup$


Firstly I should mention an example of this patter which is also given in the book that is for $n=5$ $3,2,4,1,5$ is valid whereas $3,2,5,1,4$ is not.

I have calculated that for $n=3$, the number of ways is $1+2+1=4$, for $n=4$, the number of ways is $1+3+3+1=8$, for $n=5$, the number of ways is $1+4+6+4+1=16$.

But even I can't get the general implication. Can anybody suggest me a proper way out to solve it?

Thanks for answer in advance.










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$endgroup$












  • $begingroup$
    Hint: for the $n$ odd case, what if the integers were $-frac{n-1}{2},ldots,0,ldots,frac{n-1}{2}$? Similarly, if $n$ is even, what if they were $-frac{n-2}{2},ldots,0,ldots,frac{n}{2}$? Now notice that this translation doesn't change anything.
    $endgroup$
    – user3482749
    Jan 1 at 17:55










  • $begingroup$
    Note: swapping $i$ and $n+1-i$ throughout changes a "good" sequence into another. It follows that the answer must always be even....calling that $25$ into question.
    $endgroup$
    – lulu
    Jan 1 at 18:02
















0












$begingroup$


Firstly I should mention an example of this patter which is also given in the book that is for $n=5$ $3,2,4,1,5$ is valid whereas $3,2,5,1,4$ is not.

I have calculated that for $n=3$, the number of ways is $1+2+1=4$, for $n=4$, the number of ways is $1+3+3+1=8$, for $n=5$, the number of ways is $1+4+6+4+1=16$.

But even I can't get the general implication. Can anybody suggest me a proper way out to solve it?

Thanks for answer in advance.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: for the $n$ odd case, what if the integers were $-frac{n-1}{2},ldots,0,ldots,frac{n-1}{2}$? Similarly, if $n$ is even, what if they were $-frac{n-2}{2},ldots,0,ldots,frac{n}{2}$? Now notice that this translation doesn't change anything.
    $endgroup$
    – user3482749
    Jan 1 at 17:55










  • $begingroup$
    Note: swapping $i$ and $n+1-i$ throughout changes a "good" sequence into another. It follows that the answer must always be even....calling that $25$ into question.
    $endgroup$
    – lulu
    Jan 1 at 18:02














0












0








0





$begingroup$


Firstly I should mention an example of this patter which is also given in the book that is for $n=5$ $3,2,4,1,5$ is valid whereas $3,2,5,1,4$ is not.

I have calculated that for $n=3$, the number of ways is $1+2+1=4$, for $n=4$, the number of ways is $1+3+3+1=8$, for $n=5$, the number of ways is $1+4+6+4+1=16$.

But even I can't get the general implication. Can anybody suggest me a proper way out to solve it?

Thanks for answer in advance.










share|cite|improve this question











$endgroup$




Firstly I should mention an example of this patter which is also given in the book that is for $n=5$ $3,2,4,1,5$ is valid whereas $3,2,5,1,4$ is not.

I have calculated that for $n=3$, the number of ways is $1+2+1=4$, for $n=4$, the number of ways is $1+3+3+1=8$, for $n=5$, the number of ways is $1+4+6+4+1=16$.

But even I can't get the general implication. Can anybody suggest me a proper way out to solve it?

Thanks for answer in advance.







number-theory permutations combinations






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edited Jan 1 at 18:22







Biswarup Saha

















asked Jan 1 at 17:53









Biswarup SahaBiswarup Saha

557110




557110












  • $begingroup$
    Hint: for the $n$ odd case, what if the integers were $-frac{n-1}{2},ldots,0,ldots,frac{n-1}{2}$? Similarly, if $n$ is even, what if they were $-frac{n-2}{2},ldots,0,ldots,frac{n}{2}$? Now notice that this translation doesn't change anything.
    $endgroup$
    – user3482749
    Jan 1 at 17:55










  • $begingroup$
    Note: swapping $i$ and $n+1-i$ throughout changes a "good" sequence into another. It follows that the answer must always be even....calling that $25$ into question.
    $endgroup$
    – lulu
    Jan 1 at 18:02


















  • $begingroup$
    Hint: for the $n$ odd case, what if the integers were $-frac{n-1}{2},ldots,0,ldots,frac{n-1}{2}$? Similarly, if $n$ is even, what if they were $-frac{n-2}{2},ldots,0,ldots,frac{n}{2}$? Now notice that this translation doesn't change anything.
    $endgroup$
    – user3482749
    Jan 1 at 17:55










  • $begingroup$
    Note: swapping $i$ and $n+1-i$ throughout changes a "good" sequence into another. It follows that the answer must always be even....calling that $25$ into question.
    $endgroup$
    – lulu
    Jan 1 at 18:02
















$begingroup$
Hint: for the $n$ odd case, what if the integers were $-frac{n-1}{2},ldots,0,ldots,frac{n-1}{2}$? Similarly, if $n$ is even, what if they were $-frac{n-2}{2},ldots,0,ldots,frac{n}{2}$? Now notice that this translation doesn't change anything.
$endgroup$
– user3482749
Jan 1 at 17:55




$begingroup$
Hint: for the $n$ odd case, what if the integers were $-frac{n-1}{2},ldots,0,ldots,frac{n-1}{2}$? Similarly, if $n$ is even, what if they were $-frac{n-2}{2},ldots,0,ldots,frac{n}{2}$? Now notice that this translation doesn't change anything.
$endgroup$
– user3482749
Jan 1 at 17:55












$begingroup$
Note: swapping $i$ and $n+1-i$ throughout changes a "good" sequence into another. It follows that the answer must always be even....calling that $25$ into question.
$endgroup$
– lulu
Jan 1 at 18:02




$begingroup$
Note: swapping $i$ and $n+1-i$ throughout changes a "good" sequence into another. It follows that the answer must always be even....calling that $25$ into question.
$endgroup$
– lulu
Jan 1 at 18:02










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$begingroup$

You can use induction. Let $A_n$ be the number of such choices for $1,ldots, n$. The number that must go in the last place (the $n$-th place) is either $n$ or 1. Then after the number that goes in the $n$-th place is decided, the number of ways to arrange the remaining $n-1$ numbers in the first $n-1$ places $1,2,ldots, n-1$ is $A_{n-1}$. Thus $A_n = 2A_{n-1}$.



Finish by noting $A_2 = 2$ to get $A_n=2^{n-1}$.






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    $begingroup$

    You can use induction. Let $A_n$ be the number of such choices for $1,ldots, n$. The number that must go in the last place (the $n$-th place) is either $n$ or 1. Then after the number that goes in the $n$-th place is decided, the number of ways to arrange the remaining $n-1$ numbers in the first $n-1$ places $1,2,ldots, n-1$ is $A_{n-1}$. Thus $A_n = 2A_{n-1}$.



    Finish by noting $A_2 = 2$ to get $A_n=2^{n-1}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You can use induction. Let $A_n$ be the number of such choices for $1,ldots, n$. The number that must go in the last place (the $n$-th place) is either $n$ or 1. Then after the number that goes in the $n$-th place is decided, the number of ways to arrange the remaining $n-1$ numbers in the first $n-1$ places $1,2,ldots, n-1$ is $A_{n-1}$. Thus $A_n = 2A_{n-1}$.



      Finish by noting $A_2 = 2$ to get $A_n=2^{n-1}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You can use induction. Let $A_n$ be the number of such choices for $1,ldots, n$. The number that must go in the last place (the $n$-th place) is either $n$ or 1. Then after the number that goes in the $n$-th place is decided, the number of ways to arrange the remaining $n-1$ numbers in the first $n-1$ places $1,2,ldots, n-1$ is $A_{n-1}$. Thus $A_n = 2A_{n-1}$.



        Finish by noting $A_2 = 2$ to get $A_n=2^{n-1}$.






        share|cite|improve this answer









        $endgroup$



        You can use induction. Let $A_n$ be the number of such choices for $1,ldots, n$. The number that must go in the last place (the $n$-th place) is either $n$ or 1. Then after the number that goes in the $n$-th place is decided, the number of ways to arrange the remaining $n-1$ numbers in the first $n-1$ places $1,2,ldots, n-1$ is $A_{n-1}$. Thus $A_n = 2A_{n-1}$.



        Finish by noting $A_2 = 2$ to get $A_n=2^{n-1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 17:59









        MikeMike

        3,507411




        3,507411






























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