Integers $1,2,…,n$ are placed in a way that each value is either bigger or smaller than all preceding...
$begingroup$
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.
number-theory permutations combinations
asked Jan 1 at 17:53
Biswarup SahaBiswarup Saha
557110
$endgroup$
add a comment |
$begingroup$
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.
number-theory permutations combinations
asked Jan 1 at 17:53
Biswarup SahaBiswarup Saha
557110
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
number-theory permutations combinations
asked Jan 1 at 17:53
Biswarup SahaBiswarup Saha
557110
$endgroup$
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
number-theory permutations combinations
asked Jan 1 at 17:53
Biswarup SahaBiswarup Saha
557110
asked Jan 1 at 17:53
Biswarup SahaBiswarup Saha
557110
edited Jan 1 at 18:22
Biswarup Saha
asked Jan 1 at 17:53
Biswarup SahaBiswarup Saha
557110
asked Jan 1 at 17:53
Biswarup SahaBiswarup Saha
557110
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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 $A_n=2^{n-1}$.
answered Jan 1 at 17:59
MikeMike
3,507411
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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 $A_n=2^{n-1}$.
answered Jan 1 at 17:59
MikeMike
3,507411
$endgroup$
add a comment |
$begingroup$
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 $A_n=2^{n-1}$.
answered Jan 1 at 17:59
MikeMike
3,507411
$endgroup$
add a comment |
$begingroup$
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 $A_n=2^{n-1}$.
answered Jan 1 at 17:59
MikeMike
3,507411
$endgroup$
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 $A_n=2^{n-1}$.
answered Jan 1 at 17:59
MikeMike
3,507411
answered Jan 1 at 17:59
MikeMike
3,507411
answered Jan 1 at 17:59
MikeMike
3,507411
answered Jan 1 at 17:59
MikeMike
3,507411
3,507411
add a comment |
add a comment |
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$begingroup$
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