Arrangements of MISSISSIPPI with all S's and P's separated












4












$begingroup$


If the all S,P are separated in the word MISSISSIPPI then the total possible arrangements are $kfrac {10!}{4!.4!} $ then value of k is? Now all S,P are separated so we can have slash method. Thus they can be placed alternatively in $6! $ ways and rest 5 can be placed in $5! $ ways. Thus we equate to get $$6!.5!=kfrac {10!}{4!.4!} $$ to get $k=13.7$ while $k=4/3$ .Where is my mistake Thanks






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can a S and a P occur together ?
    $endgroup$
    – Shailesh
    Sep 6 '16 at 10:03










  • $begingroup$
    Yes they can as nothing is mentioned in the question
    $endgroup$
    – Archis Welankar
    Sep 6 '16 at 10:11










  • $begingroup$
    Your book answer is incorrect. $k = 1$ is correct as pointed out by the answer by trueblueanil.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:54
















4












$begingroup$


If the all S,P are separated in the word MISSISSIPPI then the total possible arrangements are $kfrac {10!}{4!.4!} $ then value of k is? Now all S,P are separated so we can have slash method. Thus they can be placed alternatively in $6! $ ways and rest 5 can be placed in $5! $ ways. Thus we equate to get $$6!.5!=kfrac {10!}{4!.4!} $$ to get $k=13.7$ while $k=4/3$ .Where is my mistake Thanks






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can a S and a P occur together ?
    $endgroup$
    – Shailesh
    Sep 6 '16 at 10:03










  • $begingroup$
    Yes they can as nothing is mentioned in the question
    $endgroup$
    – Archis Welankar
    Sep 6 '16 at 10:11










  • $begingroup$
    Your book answer is incorrect. $k = 1$ is correct as pointed out by the answer by trueblueanil.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:54














4












4








4


3



$begingroup$


If the all S,P are separated in the word MISSISSIPPI then the total possible arrangements are $kfrac {10!}{4!.4!} $ then value of k is? Now all S,P are separated so we can have slash method. Thus they can be placed alternatively in $6! $ ways and rest 5 can be placed in $5! $ ways. Thus we equate to get $$6!.5!=kfrac {10!}{4!.4!} $$ to get $k=13.7$ while $k=4/3$ .Where is my mistake Thanks






combinatorics







share|cite|improve this question











$endgroup$




If the all S,P are separated in the word MISSISSIPPI then the total possible arrangements are $kfrac {10!}{4!.4!} $ then value of k is? Now all S,P are separated so we can have slash method. Thus they can be placed alternatively in $6! $ ways and rest 5 can be placed in $5! $ ways. Thus we equate to get $$6!.5!=kfrac {10!}{4!.4!} $$ to get $k=13.7$ while $k=4/3$ .Where is my mistake Thanks






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 29 '16 at 14:35









Shailesh

3,99692134




3,99692134










asked Sep 6 '16 at 9:52









Archis WelankarArchis Welankar

12.1k41641




12.1k41641












  • $begingroup$
    Can a S and a P occur together ?
    $endgroup$
    – Shailesh
    Sep 6 '16 at 10:03










  • $begingroup$
    Yes they can as nothing is mentioned in the question
    $endgroup$
    – Archis Welankar
    Sep 6 '16 at 10:11










  • $begingroup$
    Your book answer is incorrect. $k = 1$ is correct as pointed out by the answer by trueblueanil.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:54


















  • $begingroup$
    Can a S and a P occur together ?
    $endgroup$
    – Shailesh
    Sep 6 '16 at 10:03










  • $begingroup$
    Yes they can as nothing is mentioned in the question
    $endgroup$
    – Archis Welankar
    Sep 6 '16 at 10:11










  • $begingroup$
    Your book answer is incorrect. $k = 1$ is correct as pointed out by the answer by trueblueanil.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:54
















$begingroup$
Can a S and a P occur together ?
$endgroup$
– Shailesh
Sep 6 '16 at 10:03




$begingroup$
Can a S and a P occur together ?
$endgroup$
– Shailesh
Sep 6 '16 at 10:03












$begingroup$
Yes they can as nothing is mentioned in the question
$endgroup$
– Archis Welankar
Sep 6 '16 at 10:11




$begingroup$
Yes they can as nothing is mentioned in the question
$endgroup$
– Archis Welankar
Sep 6 '16 at 10:11












$begingroup$
Your book answer is incorrect. $k = 1$ is correct as pointed out by the answer by trueblueanil.
$endgroup$
– Shailesh
Sep 7 '16 at 5:54




$begingroup$
Your book answer is incorrect. $k = 1$ is correct as pointed out by the answer by trueblueanil.
$endgroup$
– Shailesh
Sep 7 '16 at 5:54










1 Answer
1






active

oldest

votes


















5












$begingroup$

Leaving out the $S's$ for the moment, there are $frac{7!}{4!2!} = 105$ permutations of $PPMIIII$



Of these $frac{6!}{4!} = 30$ will have the $P's$ together, thus $75$ will have the $P's$ apart.



For the together cases, we need to borrow an $S$ to separate them, e.g. $boxed{PSP}MIIII$



The remaining $3; S's$ can be inserted in the gaps between units (including ends) in $binom73$ ways



For cases with $P's$ already separate, e.g. PMPIIII, the $4; S's$ can be inserted similarly in $binom84$ ways.



Thus permissible permutations $= 30binom73 + 75binom84 = 6300$



For better or worse, this give a value of $k=1$




Use of heavy artillery



To make assurance doubly sure, I have used Jair Taylor's formula to confirm the answer.



Define polynomials for $kgeq 1$ by $q_k(x) =
sum_{i=1}^k frac{(-1)^{i-k}}{i!} {k-1 choose i-1}x^i$.



e.g. for $k=2, q_2(x)$ works out to ${(x^2-2x)}/2!$



For the nonce, we shall take the $I's$ to be distinct, divide by $4!$ at the end of the calculations. So we have five distinct alphabets, one of length $2$ and one of length $4$



The number of permutations will be given by




$$int_0^infty prod_j q_{k_j}(x), e^{-x},dx.$$




Wolframalpha gives the answer as $151,200$



Dividing by $4!$ to take care of the $4 I's$, we again get the answer of $6300$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your argument is flawless - not checked the 'heavy artillery' one. Also one Aditya Raut confirmed 6300 with Python code. +1 for all the efforts.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:52












  • $begingroup$
    Im just a beginner so I didnt see you computer techniques but the first part os awesom thanks
    $endgroup$
    – Archis Welankar
    Sep 7 '16 at 10:49










  • $begingroup$
    @ArchisWelankar Then upvote it too !
    $endgroup$
    – Shailesh
    Sep 7 '16 at 11:35










  • $begingroup$
    You're welcome !
    $endgroup$
    – true blue anil
    Sep 7 '16 at 14:30











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









5












$begingroup$

Leaving out the $S's$ for the moment, there are $frac{7!}{4!2!} = 105$ permutations of $PPMIIII$



Of these $frac{6!}{4!} = 30$ will have the $P's$ together, thus $75$ will have the $P's$ apart.



For the together cases, we need to borrow an $S$ to separate them, e.g. $boxed{PSP}MIIII$



The remaining $3; S's$ can be inserted in the gaps between units (including ends) in $binom73$ ways



For cases with $P's$ already separate, e.g. PMPIIII, the $4; S's$ can be inserted similarly in $binom84$ ways.



Thus permissible permutations $= 30binom73 + 75binom84 = 6300$



For better or worse, this give a value of $k=1$




Use of heavy artillery



To make assurance doubly sure, I have used Jair Taylor's formula to confirm the answer.



Define polynomials for $kgeq 1$ by $q_k(x) =
sum_{i=1}^k frac{(-1)^{i-k}}{i!} {k-1 choose i-1}x^i$.



e.g. for $k=2, q_2(x)$ works out to ${(x^2-2x)}/2!$



For the nonce, we shall take the $I's$ to be distinct, divide by $4!$ at the end of the calculations. So we have five distinct alphabets, one of length $2$ and one of length $4$



The number of permutations will be given by




$$int_0^infty prod_j q_{k_j}(x), e^{-x},dx.$$




Wolframalpha gives the answer as $151,200$



Dividing by $4!$ to take care of the $4 I's$, we again get the answer of $6300$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your argument is flawless - not checked the 'heavy artillery' one. Also one Aditya Raut confirmed 6300 with Python code. +1 for all the efforts.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:52












  • $begingroup$
    Im just a beginner so I didnt see you computer techniques but the first part os awesom thanks
    $endgroup$
    – Archis Welankar
    Sep 7 '16 at 10:49










  • $begingroup$
    @ArchisWelankar Then upvote it too !
    $endgroup$
    – Shailesh
    Sep 7 '16 at 11:35










  • $begingroup$
    You're welcome !
    $endgroup$
    – true blue anil
    Sep 7 '16 at 14:30
















5












$begingroup$

Leaving out the $S's$ for the moment, there are $frac{7!}{4!2!} = 105$ permutations of $PPMIIII$



Of these $frac{6!}{4!} = 30$ will have the $P's$ together, thus $75$ will have the $P's$ apart.



For the together cases, we need to borrow an $S$ to separate them, e.g. $boxed{PSP}MIIII$



The remaining $3; S's$ can be inserted in the gaps between units (including ends) in $binom73$ ways



For cases with $P's$ already separate, e.g. PMPIIII, the $4; S's$ can be inserted similarly in $binom84$ ways.



Thus permissible permutations $= 30binom73 + 75binom84 = 6300$



For better or worse, this give a value of $k=1$




Use of heavy artillery



To make assurance doubly sure, I have used Jair Taylor's formula to confirm the answer.



Define polynomials for $kgeq 1$ by $q_k(x) =
sum_{i=1}^k frac{(-1)^{i-k}}{i!} {k-1 choose i-1}x^i$.



e.g. for $k=2, q_2(x)$ works out to ${(x^2-2x)}/2!$



For the nonce, we shall take the $I's$ to be distinct, divide by $4!$ at the end of the calculations. So we have five distinct alphabets, one of length $2$ and one of length $4$



The number of permutations will be given by




$$int_0^infty prod_j q_{k_j}(x), e^{-x},dx.$$




Wolframalpha gives the answer as $151,200$



Dividing by $4!$ to take care of the $4 I's$, we again get the answer of $6300$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your argument is flawless - not checked the 'heavy artillery' one. Also one Aditya Raut confirmed 6300 with Python code. +1 for all the efforts.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:52












  • $begingroup$
    Im just a beginner so I didnt see you computer techniques but the first part os awesom thanks
    $endgroup$
    – Archis Welankar
    Sep 7 '16 at 10:49










  • $begingroup$
    @ArchisWelankar Then upvote it too !
    $endgroup$
    – Shailesh
    Sep 7 '16 at 11:35










  • $begingroup$
    You're welcome !
    $endgroup$
    – true blue anil
    Sep 7 '16 at 14:30














5












5








5





$begingroup$

Leaving out the $S's$ for the moment, there are $frac{7!}{4!2!} = 105$ permutations of $PPMIIII$



Of these $frac{6!}{4!} = 30$ will have the $P's$ together, thus $75$ will have the $P's$ apart.



For the together cases, we need to borrow an $S$ to separate them, e.g. $boxed{PSP}MIIII$



The remaining $3; S's$ can be inserted in the gaps between units (including ends) in $binom73$ ways



For cases with $P's$ already separate, e.g. PMPIIII, the $4; S's$ can be inserted similarly in $binom84$ ways.



Thus permissible permutations $= 30binom73 + 75binom84 = 6300$



For better or worse, this give a value of $k=1$




Use of heavy artillery



To make assurance doubly sure, I have used Jair Taylor's formula to confirm the answer.



Define polynomials for $kgeq 1$ by $q_k(x) =
sum_{i=1}^k frac{(-1)^{i-k}}{i!} {k-1 choose i-1}x^i$.



e.g. for $k=2, q_2(x)$ works out to ${(x^2-2x)}/2!$



For the nonce, we shall take the $I's$ to be distinct, divide by $4!$ at the end of the calculations. So we have five distinct alphabets, one of length $2$ and one of length $4$



The number of permutations will be given by




$$int_0^infty prod_j q_{k_j}(x), e^{-x},dx.$$




Wolframalpha gives the answer as $151,200$



Dividing by $4!$ to take care of the $4 I's$, we again get the answer of $6300$






share|cite|improve this answer











$endgroup$



Leaving out the $S's$ for the moment, there are $frac{7!}{4!2!} = 105$ permutations of $PPMIIII$



Of these $frac{6!}{4!} = 30$ will have the $P's$ together, thus $75$ will have the $P's$ apart.



For the together cases, we need to borrow an $S$ to separate them, e.g. $boxed{PSP}MIIII$



The remaining $3; S's$ can be inserted in the gaps between units (including ends) in $binom73$ ways



For cases with $P's$ already separate, e.g. PMPIIII, the $4; S's$ can be inserted similarly in $binom84$ ways.



Thus permissible permutations $= 30binom73 + 75binom84 = 6300$



For better or worse, this give a value of $k=1$




Use of heavy artillery



To make assurance doubly sure, I have used Jair Taylor's formula to confirm the answer.



Define polynomials for $kgeq 1$ by $q_k(x) =
sum_{i=1}^k frac{(-1)^{i-k}}{i!} {k-1 choose i-1}x^i$.



e.g. for $k=2, q_2(x)$ works out to ${(x^2-2x)}/2!$



For the nonce, we shall take the $I's$ to be distinct, divide by $4!$ at the end of the calculations. So we have five distinct alphabets, one of length $2$ and one of length $4$



The number of permutations will be given by




$$int_0^infty prod_j q_{k_j}(x), e^{-x},dx.$$




Wolframalpha gives the answer as $151,200$



Dividing by $4!$ to take care of the $4 I's$, we again get the answer of $6300$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '18 at 12:01









amWhy

1




1










answered Sep 6 '16 at 12:07









true blue aniltrue blue anil

20.5k11941




20.5k11941












  • $begingroup$
    Your argument is flawless - not checked the 'heavy artillery' one. Also one Aditya Raut confirmed 6300 with Python code. +1 for all the efforts.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:52












  • $begingroup$
    Im just a beginner so I didnt see you computer techniques but the first part os awesom thanks
    $endgroup$
    – Archis Welankar
    Sep 7 '16 at 10:49










  • $begingroup$
    @ArchisWelankar Then upvote it too !
    $endgroup$
    – Shailesh
    Sep 7 '16 at 11:35










  • $begingroup$
    You're welcome !
    $endgroup$
    – true blue anil
    Sep 7 '16 at 14:30


















  • $begingroup$
    Your argument is flawless - not checked the 'heavy artillery' one. Also one Aditya Raut confirmed 6300 with Python code. +1 for all the efforts.
    $endgroup$
    – Shailesh
    Sep 7 '16 at 5:52












  • $begingroup$
    Im just a beginner so I didnt see you computer techniques but the first part os awesom thanks
    $endgroup$
    – Archis Welankar
    Sep 7 '16 at 10:49










  • $begingroup$
    @ArchisWelankar Then upvote it too !
    $endgroup$
    – Shailesh
    Sep 7 '16 at 11:35










  • $begingroup$
    You're welcome !
    $endgroup$
    – true blue anil
    Sep 7 '16 at 14:30
















$begingroup$
Your argument is flawless - not checked the 'heavy artillery' one. Also one Aditya Raut confirmed 6300 with Python code. +1 for all the efforts.
$endgroup$
– Shailesh
Sep 7 '16 at 5:52






$begingroup$
Your argument is flawless - not checked the 'heavy artillery' one. Also one Aditya Raut confirmed 6300 with Python code. +1 for all the efforts.
$endgroup$
– Shailesh
Sep 7 '16 at 5:52














$begingroup$
Im just a beginner so I didnt see you computer techniques but the first part os awesom thanks
$endgroup$
– Archis Welankar
Sep 7 '16 at 10:49




$begingroup$
Im just a beginner so I didnt see you computer techniques but the first part os awesom thanks
$endgroup$
– Archis Welankar
Sep 7 '16 at 10:49












$begingroup$
@ArchisWelankar Then upvote it too !
$endgroup$
– Shailesh
Sep 7 '16 at 11:35




$begingroup$
@ArchisWelankar Then upvote it too !
$endgroup$
– Shailesh
Sep 7 '16 at 11:35












$begingroup$
You're welcome !
$endgroup$
– true blue anil
Sep 7 '16 at 14:30




$begingroup$
You're welcome !
$endgroup$
– true blue anil
Sep 7 '16 at 14:30


















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