Letters in Mailboxes so that none are in the right one












1












$begingroup$


In how many ways can 5 letters be put in 5 mailboxes such that none are placed in the right one ? I creatively thought labeling the letters & mailboxes 1-5 , it would be saying letter 1 can be placed in any of the other four . So a total of 4^5 ways .this was not one of the answers . Then i thought hmm lets write as an example 13254 as a sequence above 12345 ..this sequence is a ' throw-out ' case because the 1 cant go with the 1. This would generate _ _ a two digit number ..one for the letter and one for the mailbox .as an example 12 is legit . ( letter 1 in mailbox 2 ) while the pairs 11,22, ect are not .
This would give 5 possabilities for the first number and 5 for the second , or 5x5 =25. But then we must subtract out the posabilities of the 11, 22, 33, ect
This gives 25- 5 =20 but that is STILL not one of the answers






combinatorics discrete-mathematics







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$endgroup$








  • 1




    $begingroup$
    Permutations in which no object is left in its proper place are called derangements.
    $endgroup$
    – N. F. Taussig
    Dec 25 '18 at 22:02










  • $begingroup$
    This is called $D_5,$ the number of derangements of $5$ things.
    $endgroup$
    – coffeemath
    Dec 25 '18 at 22:03
















1












$begingroup$


In how many ways can 5 letters be put in 5 mailboxes such that none are placed in the right one ? I creatively thought labeling the letters & mailboxes 1-5 , it would be saying letter 1 can be placed in any of the other four . So a total of 4^5 ways .this was not one of the answers . Then i thought hmm lets write as an example 13254 as a sequence above 12345 ..this sequence is a ' throw-out ' case because the 1 cant go with the 1. This would generate _ _ a two digit number ..one for the letter and one for the mailbox .as an example 12 is legit . ( letter 1 in mailbox 2 ) while the pairs 11,22, ect are not .
This would give 5 possabilities for the first number and 5 for the second , or 5x5 =25. But then we must subtract out the posabilities of the 11, 22, 33, ect
This gives 25- 5 =20 but that is STILL not one of the answers






combinatorics discrete-mathematics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Permutations in which no object is left in its proper place are called derangements.
    $endgroup$
    – N. F. Taussig
    Dec 25 '18 at 22:02










  • $begingroup$
    This is called $D_5,$ the number of derangements of $5$ things.
    $endgroup$
    – coffeemath
    Dec 25 '18 at 22:03














1












1








1





$begingroup$


In how many ways can 5 letters be put in 5 mailboxes such that none are placed in the right one ? I creatively thought labeling the letters & mailboxes 1-5 , it would be saying letter 1 can be placed in any of the other four . So a total of 4^5 ways .this was not one of the answers . Then i thought hmm lets write as an example 13254 as a sequence above 12345 ..this sequence is a ' throw-out ' case because the 1 cant go with the 1. This would generate _ _ a two digit number ..one for the letter and one for the mailbox .as an example 12 is legit . ( letter 1 in mailbox 2 ) while the pairs 11,22, ect are not .
This would give 5 possabilities for the first number and 5 for the second , or 5x5 =25. But then we must subtract out the posabilities of the 11, 22, 33, ect
This gives 25- 5 =20 but that is STILL not one of the answers






combinatorics discrete-mathematics







share|cite|improve this question











$endgroup$




In how many ways can 5 letters be put in 5 mailboxes such that none are placed in the right one ? I creatively thought labeling the letters & mailboxes 1-5 , it would be saying letter 1 can be placed in any of the other four . So a total of 4^5 ways .this was not one of the answers . Then i thought hmm lets write as an example 13254 as a sequence above 12345 ..this sequence is a ' throw-out ' case because the 1 cant go with the 1. This would generate _ _ a two digit number ..one for the letter and one for the mailbox .as an example 12 is legit . ( letter 1 in mailbox 2 ) while the pairs 11,22, ect are not .
This would give 5 possabilities for the first number and 5 for the second , or 5x5 =25. But then we must subtract out the posabilities of the 11, 22, 33, ect
This gives 25- 5 =20 but that is STILL not one of the answers






combinatorics discrete-mathematics




combinatorics discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 11 at 11:43









Maria Mazur

50k1361125




50k1361125










asked Dec 25 '18 at 21:59









RandinRandin

347116




347116








  • 1




    $begingroup$
    Permutations in which no object is left in its proper place are called derangements.
    $endgroup$
    – N. F. Taussig
    Dec 25 '18 at 22:02










  • $begingroup$
    This is called $D_5,$ the number of derangements of $5$ things.
    $endgroup$
    – coffeemath
    Dec 25 '18 at 22:03














  • 1




    $begingroup$
    Permutations in which no object is left in its proper place are called derangements.
    $endgroup$
    – N. F. Taussig
    Dec 25 '18 at 22:02










  • $begingroup$
    This is called $D_5,$ the number of derangements of $5$ things.
    $endgroup$
    – coffeemath
    Dec 25 '18 at 22:03








1




1




$begingroup$
Permutations in which no object is left in its proper place are called derangements.
$endgroup$
– N. F. Taussig
Dec 25 '18 at 22:02




$begingroup$
Permutations in which no object is left in its proper place are called derangements.
$endgroup$
– N. F. Taussig
Dec 25 '18 at 22:02












$begingroup$
This is called $D_5,$ the number of derangements of $5$ things.
$endgroup$
– coffeemath
Dec 25 '18 at 22:03




$begingroup$
This is called $D_5,$ the number of derangements of $5$ things.
$endgroup$
– coffeemath
Dec 25 '18 at 22:03










2 Answers
2






active

oldest

votes


















1












$begingroup$

These are known as derangements.



Look here:
https://en.wikipedia.org/wiki/Derangement






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You all are derranged .😆😂🤣
    $endgroup$
    – Randin
    Dec 26 '18 at 6:16










  • $begingroup$
    I dont see why my way that seemed so intuitively true was errored ? 😣 btw merry x^3 mas fellow math nerds 🤓🎄🎄🎄👽📯🎶🎁🎎
    $endgroup$
    – Randin
    Dec 26 '18 at 6:17



















2












$begingroup$

So you count bijective functions from $left{1,2,3,4,5right}$ to $left{1,2,3,4,5right}$ such that $f(i) ne i$



I would do like this: Let $A_i$ be a set of functions with $f(i)=i$. Then



$|A_i|= 4!$ and



$|A_icap A_j|=3!$ and



$|A_icap A_jcap A_k|=2!$ and



$|A_icap A_jcap A_kcap A_n|=1$



You are interested in $|A_1'cup A_2'...cup A_5'|$. Let's use PIE



$$|A_1'cup A_2'...cup A_5'| = 5!-5cdot 4!+{5choose 2}3!-{5choose 3}2!+ {5choose 4}1!- {5choose 5}cdot 1=...$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    NB using the definition of factorial, we can rewrite the sum as $5! (frac{1}{0!} - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!})$. Then using the power series for $exp$ and a straightforward estimate shows that we can write this as $left[frac{5!}{e}right]$ (here $[a]$ denotes the integer closest to $a$) and that this formula generalizes to when we replace $5$ with any positive integer.
    $endgroup$
    – Travis
    Dec 25 '18 at 22:38














Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

These are known as derangements.



Look here:
https://en.wikipedia.org/wiki/Derangement






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You all are derranged .😆😂🤣
    $endgroup$
    – Randin
    Dec 26 '18 at 6:16










  • $begingroup$
    I dont see why my way that seemed so intuitively true was errored ? 😣 btw merry x^3 mas fellow math nerds 🤓🎄🎄🎄👽📯🎶🎁🎎
    $endgroup$
    – Randin
    Dec 26 '18 at 6:17
















1












$begingroup$

These are known as derangements.



Look here:
https://en.wikipedia.org/wiki/Derangement






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You all are derranged .😆😂🤣
    $endgroup$
    – Randin
    Dec 26 '18 at 6:16










  • $begingroup$
    I dont see why my way that seemed so intuitively true was errored ? 😣 btw merry x^3 mas fellow math nerds 🤓🎄🎄🎄👽📯🎶🎁🎎
    $endgroup$
    – Randin
    Dec 26 '18 at 6:17














1












1








1





$begingroup$

These are known as derangements.



Look here:
https://en.wikipedia.org/wiki/Derangement






share|cite|improve this answer









$endgroup$



These are known as derangements.



Look here:
https://en.wikipedia.org/wiki/Derangement







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 22:02









marty cohenmarty cohen

75.1k549130




75.1k549130












  • $begingroup$
    You all are derranged .😆😂🤣
    $endgroup$
    – Randin
    Dec 26 '18 at 6:16










  • $begingroup$
    I dont see why my way that seemed so intuitively true was errored ? 😣 btw merry x^3 mas fellow math nerds 🤓🎄🎄🎄👽📯🎶🎁🎎
    $endgroup$
    – Randin
    Dec 26 '18 at 6:17


















  • $begingroup$
    You all are derranged .😆😂🤣
    $endgroup$
    – Randin
    Dec 26 '18 at 6:16










  • $begingroup$
    I dont see why my way that seemed so intuitively true was errored ? 😣 btw merry x^3 mas fellow math nerds 🤓🎄🎄🎄👽📯🎶🎁🎎
    $endgroup$
    – Randin
    Dec 26 '18 at 6:17
















$begingroup$
You all are derranged .😆😂🤣
$endgroup$
– Randin
Dec 26 '18 at 6:16




$begingroup$
You all are derranged .😆😂🤣
$endgroup$
– Randin
Dec 26 '18 at 6:16












$begingroup$
I dont see why my way that seemed so intuitively true was errored ? 😣 btw merry x^3 mas fellow math nerds 🤓🎄🎄🎄👽📯🎶🎁🎎
$endgroup$
– Randin
Dec 26 '18 at 6:17




$begingroup$
I dont see why my way that seemed so intuitively true was errored ? 😣 btw merry x^3 mas fellow math nerds 🤓🎄🎄🎄👽📯🎶🎁🎎
$endgroup$
– Randin
Dec 26 '18 at 6:17











2












$begingroup$

So you count bijective functions from $left{1,2,3,4,5right}$ to $left{1,2,3,4,5right}$ such that $f(i) ne i$



I would do like this: Let $A_i$ be a set of functions with $f(i)=i$. Then



$|A_i|= 4!$ and



$|A_icap A_j|=3!$ and



$|A_icap A_jcap A_k|=2!$ and



$|A_icap A_jcap A_kcap A_n|=1$



You are interested in $|A_1'cup A_2'...cup A_5'|$. Let's use PIE



$$|A_1'cup A_2'...cup A_5'| = 5!-5cdot 4!+{5choose 2}3!-{5choose 3}2!+ {5choose 4}1!- {5choose 5}cdot 1=...$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    NB using the definition of factorial, we can rewrite the sum as $5! (frac{1}{0!} - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!})$. Then using the power series for $exp$ and a straightforward estimate shows that we can write this as $left[frac{5!}{e}right]$ (here $[a]$ denotes the integer closest to $a$) and that this formula generalizes to when we replace $5$ with any positive integer.
    $endgroup$
    – Travis
    Dec 25 '18 at 22:38


















2












$begingroup$

So you count bijective functions from $left{1,2,3,4,5right}$ to $left{1,2,3,4,5right}$ such that $f(i) ne i$



I would do like this: Let $A_i$ be a set of functions with $f(i)=i$. Then



$|A_i|= 4!$ and



$|A_icap A_j|=3!$ and



$|A_icap A_jcap A_k|=2!$ and



$|A_icap A_jcap A_kcap A_n|=1$



You are interested in $|A_1'cup A_2'...cup A_5'|$. Let's use PIE



$$|A_1'cup A_2'...cup A_5'| = 5!-5cdot 4!+{5choose 2}3!-{5choose 3}2!+ {5choose 4}1!- {5choose 5}cdot 1=...$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    NB using the definition of factorial, we can rewrite the sum as $5! (frac{1}{0!} - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!})$. Then using the power series for $exp$ and a straightforward estimate shows that we can write this as $left[frac{5!}{e}right]$ (here $[a]$ denotes the integer closest to $a$) and that this formula generalizes to when we replace $5$ with any positive integer.
    $endgroup$
    – Travis
    Dec 25 '18 at 22:38
















2












2








2





$begingroup$

So you count bijective functions from $left{1,2,3,4,5right}$ to $left{1,2,3,4,5right}$ such that $f(i) ne i$



I would do like this: Let $A_i$ be a set of functions with $f(i)=i$. Then



$|A_i|= 4!$ and



$|A_icap A_j|=3!$ and



$|A_icap A_jcap A_k|=2!$ and



$|A_icap A_jcap A_kcap A_n|=1$



You are interested in $|A_1'cup A_2'...cup A_5'|$. Let's use PIE



$$|A_1'cup A_2'...cup A_5'| = 5!-5cdot 4!+{5choose 2}3!-{5choose 3}2!+ {5choose 4}1!- {5choose 5}cdot 1=...$$






share|cite|improve this answer









$endgroup$



So you count bijective functions from $left{1,2,3,4,5right}$ to $left{1,2,3,4,5right}$ such that $f(i) ne i$



I would do like this: Let $A_i$ be a set of functions with $f(i)=i$. Then



$|A_i|= 4!$ and



$|A_icap A_j|=3!$ and



$|A_icap A_jcap A_k|=2!$ and



$|A_icap A_jcap A_kcap A_n|=1$



You are interested in $|A_1'cup A_2'...cup A_5'|$. Let's use PIE



$$|A_1'cup A_2'...cup A_5'| = 5!-5cdot 4!+{5choose 2}3!-{5choose 3}2!+ {5choose 4}1!- {5choose 5}cdot 1=...$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 22:18









Maria MazurMaria Mazur

50k1361125




50k1361125








  • 1




    $begingroup$
    NB using the definition of factorial, we can rewrite the sum as $5! (frac{1}{0!} - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!})$. Then using the power series for $exp$ and a straightforward estimate shows that we can write this as $left[frac{5!}{e}right]$ (here $[a]$ denotes the integer closest to $a$) and that this formula generalizes to when we replace $5$ with any positive integer.
    $endgroup$
    – Travis
    Dec 25 '18 at 22:38
















  • 1




    $begingroup$
    NB using the definition of factorial, we can rewrite the sum as $5! (frac{1}{0!} - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!})$. Then using the power series for $exp$ and a straightforward estimate shows that we can write this as $left[frac{5!}{e}right]$ (here $[a]$ denotes the integer closest to $a$) and that this formula generalizes to when we replace $5$ with any positive integer.
    $endgroup$
    – Travis
    Dec 25 '18 at 22:38










1




1




$begingroup$
NB using the definition of factorial, we can rewrite the sum as $5! (frac{1}{0!} - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!})$. Then using the power series for $exp$ and a straightforward estimate shows that we can write this as $left[frac{5!}{e}right]$ (here $[a]$ denotes the integer closest to $a$) and that this formula generalizes to when we replace $5$ with any positive integer.
$endgroup$
– Travis
Dec 25 '18 at 22:38






$begingroup$
NB using the definition of factorial, we can rewrite the sum as $5! (frac{1}{0!} - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!})$. Then using the power series for $exp$ and a straightforward estimate shows that we can write this as $left[frac{5!}{e}right]$ (here $[a]$ denotes the integer closest to $a$) and that this formula generalizes to when we replace $5$ with any positive integer.
$endgroup$
– Travis
Dec 25 '18 at 22:38




















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