Ways to arrange a consecutive string of letters with no letter in the alphabetically correct placeProbability 3 eventsPermutations with fixed pointsNumber of ordered pairs of given stringHow many unique ways are there to arrange the letters in the word HATTER?How many strings of length n can be constructed with 'a', 'b', and 'c', such that there is a maximum of one 'b' and two consecutive 'c's?Combination for 3 seats from N peoplesWays to arrange letters and digits, if the letters must appear in groupsWhen and when not to consider the order of arrangement in a probability problem?No. of strings between 2 given strings in lexicographical orderHow many ways can we arrange the letters F, G, H, I and J such that three consecutive letters are in alphabetical order?
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Ways to arrange a consecutive string of letters with no letter in the alphabetically correct place
Probability 3 eventsPermutations with fixed pointsNumber of ordered pairs of given stringHow many unique ways are there to arrange the letters in the word HATTER?How many strings of length n can be constructed with 'a', 'b', and 'c', such that there is a maximum of one 'b' and two consecutive 'c's?Combination for 3 seats from N peoplesWays to arrange letters and digits, if the letters must appear in groupsWhen and when not to consider the order of arrangement in a probability problem?No. of strings between 2 given strings in lexicographical orderHow many ways can we arrange the letters F, G, H, I and J such that three consecutive letters are in alphabetical order?
$begingroup$
I am trying to figure out how to calculate the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place. For example, if we take $ABC$, then there are 6 possible arrangements:
$$ABC, ACB, BAC, BCA, CAB, CBA$$
but only in the last two none of the letters is in the alphabetically correct place (for example, $BAC$ does not satisfy the requirement, since $C$ is a third letter of the alphabet and it is in the third spot in $BAC$)
If we take $ABCD$ there are $4!=24$ possible arrangements, but by inspection one can see that only $9$ of the arrangements satisfy the given requirement.
My question is: what's the formula for calculating the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place and how to deduce such a formula?
In fact, I am also trying to understand the following: why does the ratio of all the possible arrangements to the arrangements satisfying the extra condition converge to the famous Euler's number $e$?
As always, any help will be greatly appreciated.
probability combinatorics limits combinations
asked Sep 11 '18 at 13:53
PawelPawel
3,2841022
$endgroup$
add a comment |
$begingroup$
I am trying to figure out how to calculate the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place. For example, if we take $ABC$, then there are 6 possible arrangements:
$$ABC, ACB, BAC, BCA, CAB, CBA$$
but only in the last two none of the letters is in the alphabetically correct place (for example, $BAC$ does not satisfy the requirement, since $C$ is a third letter of the alphabet and it is in the third spot in $BAC$)
If we take $ABCD$ there are $4!=24$ possible arrangements, but by inspection one can see that only $9$ of the arrangements satisfy the given requirement.
My question is: what's the formula for calculating the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place and how to deduce such a formula?
In fact, I am also trying to understand the following: why does the ratio of all the possible arrangements to the arrangements satisfying the extra condition converge to the famous Euler's number $e$?
As always, any help will be greatly appreciated.
probability combinatorics limits combinations
asked Sep 11 '18 at 13:53
PawelPawel
3,2841022
$endgroup$
7
$begingroup$
It seems that you're after the number of derangements.
$endgroup$
– José Carlos Santos
Sep 11 '18 at 13:54
add a comment |
$begingroup$
I am trying to figure out how to calculate the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place. For example, if we take $ABC$, then there are 6 possible arrangements:
$$ABC, ACB, BAC, BCA, CAB, CBA$$
but only in the last two none of the letters is in the alphabetically correct place (for example, $BAC$ does not satisfy the requirement, since $C$ is a third letter of the alphabet and it is in the third spot in $BAC$)
If we take $ABCD$ there are $4!=24$ possible arrangements, but by inspection one can see that only $9$ of the arrangements satisfy the given requirement.
My question is: what's the formula for calculating the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place and how to deduce such a formula?
In fact, I am also trying to understand the following: why does the ratio of all the possible arrangements to the arrangements satisfying the extra condition converge to the famous Euler's number $e$?
As always, any help will be greatly appreciated.
probability combinatorics limits combinations
asked Sep 11 '18 at 13:53
PawelPawel
3,2841022
$endgroup$
I am trying to figure out how to calculate the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place. For example, if we take $ABC$, then there are 6 possible arrangements:
$$ABC, ACB, BAC, BCA, CAB, CBA$$
but only in the last two none of the letters is in the alphabetically correct place (for example, $BAC$ does not satisfy the requirement, since $C$ is a third letter of the alphabet and it is in the third spot in $BAC$)
If we take $ABCD$ there are $4!=24$ possible arrangements, but by inspection one can see that only $9$ of the arrangements satisfy the given requirement.
My question is: what's the formula for calculating the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place and how to deduce such a formula?
In fact, I am also trying to understand the following: why does the ratio of all the possible arrangements to the arrangements satisfying the extra condition converge to the famous Euler's number $e$?
As always, any help will be greatly appreciated.
probability combinatorics limits combinations
probability combinatorics limits combinations
asked Sep 11 '18 at 13:53
PawelPawel
3,2841022
asked Sep 11 '18 at 13:53
PawelPawel
3,2841022
asked Sep 11 '18 at 13:53
PawelPawel
3,2841022
asked Sep 11 '18 at 13:53
PawelPawel
3,2841022
asked Sep 11 '18 at 13:53
PawelPawel
3,2841022
3,2841022
7
$begingroup$
It seems that you're after the number of derangements.
$endgroup$
– José Carlos Santos
Sep 11 '18 at 13:54
add a comment |
7
$begingroup$
It seems that you're after the number of derangements.
$endgroup$
– José Carlos Santos
Sep 11 '18 at 13:54
7
7
$begingroup$
It seems that you're after the number of derangements.
$endgroup$
– José Carlos Santos
Sep 11 '18 at 13:54
$begingroup$
It seems that you're after the number of derangements.
$endgroup$
– José Carlos Santos
Sep 11 '18 at 13:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just as Jose Carlos Santos posted in the comments, this can be solved by counting the number of derangements.
In particular a useful formula would be:
$$!n = left[ fracn!e right]$$
where $[x]$ is the nearest integer function. So the probability you are looking for would be:
$$!n =frac1n! left[ fracn!e right] rightarrow frac1e$$
answering your second question
answered Mar 29 at 15:30
Fer BelloraFer Bellora
1355
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Just as Jose Carlos Santos posted in the comments, this can be solved by counting the number of derangements.
In particular a useful formula would be:
$$!n = left[ fracn!e right]$$
where $[x]$ is the nearest integer function. So the probability you are looking for would be:
$$!n =frac1n! left[ fracn!e right] rightarrow frac1e$$
answering your second question
answered Mar 29 at 15:30
Fer BelloraFer Bellora
1355
$endgroup$
add a comment |
$begingroup$
Just as Jose Carlos Santos posted in the comments, this can be solved by counting the number of derangements.
In particular a useful formula would be:
$$!n = left[ fracn!e right]$$
where $[x]$ is the nearest integer function. So the probability you are looking for would be:
$$!n =frac1n! left[ fracn!e right] rightarrow frac1e$$
answering your second question
answered Mar 29 at 15:30
Fer BelloraFer Bellora
1355
$endgroup$
add a comment |
$begingroup$
Just as Jose Carlos Santos posted in the comments, this can be solved by counting the number of derangements.
In particular a useful formula would be:
$$!n = left[ fracn!e right]$$
where $[x]$ is the nearest integer function. So the probability you are looking for would be:
$$!n =frac1n! left[ fracn!e right] rightarrow frac1e$$
answering your second question
answered Mar 29 at 15:30
Fer BelloraFer Bellora
1355
$endgroup$
Just as Jose Carlos Santos posted in the comments, this can be solved by counting the number of derangements.
In particular a useful formula would be:
$$!n = left[ fracn!e right]$$
where $[x]$ is the nearest integer function. So the probability you are looking for would be:
$$!n =frac1n! left[ fracn!e right] rightarrow frac1e$$
answering your second question
answered Mar 29 at 15:30
Fer BelloraFer Bellora
1355
answered Mar 29 at 15:30
Fer BelloraFer Bellora
1355
answered Mar 29 at 15:30
Fer BelloraFer Bellora
1355
answered Mar 29 at 15:30
Fer BelloraFer Bellora
1355
1355
add a comment |
add a comment |
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7
$begingroup$
It seems that you're after the number of derangements.
$endgroup$
– José Carlos Santos
Sep 11 '18 at 13:54