Letter permutation MISSISISPPI, S come before any I
$begingroup$
If there is no restriction, the number of ways to organize letter of MISSISSIPPI is,
$$ frac{11!}{4!4!2!} $$
The restriction is,
all Ss come before any Is. So I group both letters, then there are total 4 letters. (PP,M, (Ss and Is)).
$$ frac{4!}{2!}$$.
Is it the correct way?
combinatorics discrete-mathematics permutations
asked Dec 8 '18 at 20:24
jayko03jayko03
1238
$endgroup$
add a comment |
$begingroup$
If there is no restriction, the number of ways to organize letter of MISSISSIPPI is,
$$ frac{11!}{4!4!2!} $$
The restriction is,
all Ss come before any Is. So I group both letters, then there are total 4 letters. (PP,M, (Ss and Is)).
$$ frac{4!}{2!}$$.
Is it the correct way?
combinatorics discrete-mathematics permutations
asked Dec 8 '18 at 20:24
jayko03jayko03
1238
$endgroup$
$begingroup$
For restriction you neglected the case like MSSPSPSIIII or SSSMSPIIPII. Think how to accomodate them also.
$endgroup$
– jayant98
Dec 8 '18 at 20:33
$begingroup$
Your idea of considering S and I the same letter is good, but it would yield (11!/8!2!1!) possibilities
$endgroup$
– josinalvo
Dec 8 '18 at 20:40
add a comment |
$begingroup$
If there is no restriction, the number of ways to organize letter of MISSISSIPPI is,
$$ frac{11!}{4!4!2!} $$
The restriction is,
all Ss come before any Is. So I group both letters, then there are total 4 letters. (PP,M, (Ss and Is)).
$$ frac{4!}{2!}$$.
Is it the correct way?
combinatorics discrete-mathematics permutations
asked Dec 8 '18 at 20:24
jayko03jayko03
1238
$endgroup$
If there is no restriction, the number of ways to organize letter of MISSISSIPPI is,
$$ frac{11!}{4!4!2!} $$
The restriction is,
all Ss come before any Is. So I group both letters, then there are total 4 letters. (PP,M, (Ss and Is)).
$$ frac{4!}{2!}$$.
Is it the correct way?
combinatorics discrete-mathematics permutations
combinatorics discrete-mathematics permutations
asked Dec 8 '18 at 20:24
jayko03jayko03
1238
asked Dec 8 '18 at 20:24
jayko03jayko03
1238
asked Dec 8 '18 at 20:24
jayko03jayko03
1238
asked Dec 8 '18 at 20:24
jayko03jayko03
1238
asked Dec 8 '18 at 20:24
jayko03jayko03
1238
1238
$begingroup$
For restriction you neglected the case like MSSPSPSIIII or SSSMSPIIPII. Think how to accomodate them also.
$endgroup$
– jayant98
Dec 8 '18 at 20:33
$begingroup$
Your idea of considering S and I the same letter is good, but it would yield (11!/8!2!1!) possibilities
$endgroup$
– josinalvo
Dec 8 '18 at 20:40
add a comment |
$begingroup$
For restriction you neglected the case like MSSPSPSIIII or SSSMSPIIPII. Think how to accomodate them also.
$endgroup$
– jayant98
Dec 8 '18 at 20:33
$begingroup$
Your idea of considering S and I the same letter is good, but it would yield (11!/8!2!1!) possibilities
$endgroup$
– josinalvo
Dec 8 '18 at 20:40
$begingroup$
For restriction you neglected the case like MSSPSPSIIII or SSSMSPIIPII. Think how to accomodate them also.
$endgroup$
– jayant98
Dec 8 '18 at 20:33
$begingroup$
For restriction you neglected the case like MSSPSPSIIII or SSSMSPIIPII. Think how to accomodate them also.
$endgroup$
– jayant98
Dec 8 '18 at 20:33
$begingroup$
Your idea of considering S and I the same letter is good, but it would yield (11!/8!2!1!) possibilities
$endgroup$
– josinalvo
Dec 8 '18 at 20:40
$begingroup$
Your idea of considering S and I the same letter is good, but it would yield (11!/8!2!1!) possibilities
$endgroup$
– josinalvo
Dec 8 '18 at 20:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, it is not. The problem merely states that every $S$ appears before any $I$, not that the $S$s have to appear together or that the $I$s have to appear together.
There are eleven letters in $MISSISSIPPI$, so we have eleven positions to fill. We can place the $M$ in eleven ways, which leaves ten open positions. We choose two of them for the $P$s. Once we have done that, there is only one way to fill the remaining positions with the $S$s and $I$s since every $S$ must appear before the first $I$. Hence, there are
$$binom{11}{1}binom{10}{2}$$
admissible arrangements.
answered Dec 8 '18 at 20:33
N. F. TaussigN. F. Taussig
44.7k103358
$endgroup$
$begingroup$
I think this is another approach. I ended up like this. First, group the Ss and Is and treat them as one letter. So there are $$ binom{11}{8} $$ way to organize for Ss and Is. Then there is rest of letters (M, P, P) $$ frac{3!}{2!} $$. Multiply these, $$ binom{11}{8} frac{3!}{2!} $$.
$endgroup$
– jayko03
Dec 10 '18 at 3:55
$begingroup$
Your count is correct. However, you should say that you are choosing eight of the eleven positions for the $S$s and $I$s (since the eight letters do not form a block). There is only way to place the $S$s and $I$s in those positions since every $S$ must appear before the first $I$. We must choose one of the remaining three positions for the M. The final two positions must be filled with the two $P$s. We get $$binom{11}{8}binom{3}{1} = binom{11}{8}frac{3!}{1!2!} = binom{11}{8}frac{3!}{2!}$$ Our approaches are equivalent. We just chose to fill the positions in different orders.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 11:45
$begingroup$
In my parenthetical remark in my previous comment, I meant to say that the eight letters may not form a block.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 16:31
add a comment |
$begingroup$
Since all S should come before I consider the word SSSSIIII. Now we need to add the letters MPP. First letter we add can take 9 positikns (anywhere between letters or at ends of SSSSIIII). Second and third can be added at 10 and 11 positions. But P appears twice so we divide by 2!. Final answer is (11!/8!)/2!
answered Dec 13 '18 at 15:51
CuriousCurious
889516
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
No, it is not. The problem merely states that every $S$ appears before any $I$, not that the $S$s have to appear together or that the $I$s have to appear together.
There are eleven letters in $MISSISSIPPI$, so we have eleven positions to fill. We can place the $M$ in eleven ways, which leaves ten open positions. We choose two of them for the $P$s. Once we have done that, there is only one way to fill the remaining positions with the $S$s and $I$s since every $S$ must appear before the first $I$. Hence, there are
$$binom{11}{1}binom{10}{2}$$
admissible arrangements.
answered Dec 8 '18 at 20:33
N. F. TaussigN. F. Taussig
44.7k103358
$endgroup$
$begingroup$
I think this is another approach. I ended up like this. First, group the Ss and Is and treat them as one letter. So there are $$ binom{11}{8} $$ way to organize for Ss and Is. Then there is rest of letters (M, P, P) $$ frac{3!}{2!} $$. Multiply these, $$ binom{11}{8} frac{3!}{2!} $$.
$endgroup$
– jayko03
Dec 10 '18 at 3:55
$begingroup$
Your count is correct. However, you should say that you are choosing eight of the eleven positions for the $S$s and $I$s (since the eight letters do not form a block). There is only way to place the $S$s and $I$s in those positions since every $S$ must appear before the first $I$. We must choose one of the remaining three positions for the M. The final two positions must be filled with the two $P$s. We get $$binom{11}{8}binom{3}{1} = binom{11}{8}frac{3!}{1!2!} = binom{11}{8}frac{3!}{2!}$$ Our approaches are equivalent. We just chose to fill the positions in different orders.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 11:45
$begingroup$
In my parenthetical remark in my previous comment, I meant to say that the eight letters may not form a block.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 16:31
add a comment |
$begingroup$
No, it is not. The problem merely states that every $S$ appears before any $I$, not that the $S$s have to appear together or that the $I$s have to appear together.
There are eleven letters in $MISSISSIPPI$, so we have eleven positions to fill. We can place the $M$ in eleven ways, which leaves ten open positions. We choose two of them for the $P$s. Once we have done that, there is only one way to fill the remaining positions with the $S$s and $I$s since every $S$ must appear before the first $I$. Hence, there are
$$binom{11}{1}binom{10}{2}$$
admissible arrangements.
answered Dec 8 '18 at 20:33
N. F. TaussigN. F. Taussig
44.7k103358
$endgroup$
$begingroup$
I think this is another approach. I ended up like this. First, group the Ss and Is and treat them as one letter. So there are $$ binom{11}{8} $$ way to organize for Ss and Is. Then there is rest of letters (M, P, P) $$ frac{3!}{2!} $$. Multiply these, $$ binom{11}{8} frac{3!}{2!} $$.
$endgroup$
– jayko03
Dec 10 '18 at 3:55
$begingroup$
Your count is correct. However, you should say that you are choosing eight of the eleven positions for the $S$s and $I$s (since the eight letters do not form a block). There is only way to place the $S$s and $I$s in those positions since every $S$ must appear before the first $I$. We must choose one of the remaining three positions for the M. The final two positions must be filled with the two $P$s. We get $$binom{11}{8}binom{3}{1} = binom{11}{8}frac{3!}{1!2!} = binom{11}{8}frac{3!}{2!}$$ Our approaches are equivalent. We just chose to fill the positions in different orders.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 11:45
$begingroup$
In my parenthetical remark in my previous comment, I meant to say that the eight letters may not form a block.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 16:31
add a comment |
$begingroup$
No, it is not. The problem merely states that every $S$ appears before any $I$, not that the $S$s have to appear together or that the $I$s have to appear together.
There are eleven letters in $MISSISSIPPI$, so we have eleven positions to fill. We can place the $M$ in eleven ways, which leaves ten open positions. We choose two of them for the $P$s. Once we have done that, there is only one way to fill the remaining positions with the $S$s and $I$s since every $S$ must appear before the first $I$. Hence, there are
$$binom{11}{1}binom{10}{2}$$
admissible arrangements.
answered Dec 8 '18 at 20:33
N. F. TaussigN. F. Taussig
44.7k103358
$endgroup$
No, it is not. The problem merely states that every $S$ appears before any $I$, not that the $S$s have to appear together or that the $I$s have to appear together.
There are eleven letters in $MISSISSIPPI$, so we have eleven positions to fill. We can place the $M$ in eleven ways, which leaves ten open positions. We choose two of them for the $P$s. Once we have done that, there is only one way to fill the remaining positions with the $S$s and $I$s since every $S$ must appear before the first $I$. Hence, there are
$$binom{11}{1}binom{10}{2}$$
admissible arrangements.
answered Dec 8 '18 at 20:33
N. F. TaussigN. F. Taussig
44.7k103358
answered Dec 8 '18 at 20:33
N. F. TaussigN. F. Taussig
44.7k103358
answered Dec 8 '18 at 20:33
N. F. TaussigN. F. Taussig
44.7k103358
answered Dec 8 '18 at 20:33
N. F. TaussigN. F. Taussig
44.7k103358
44.7k103358
$begingroup$
I think this is another approach. I ended up like this. First, group the Ss and Is and treat them as one letter. So there are $$ binom{11}{8} $$ way to organize for Ss and Is. Then there is rest of letters (M, P, P) $$ frac{3!}{2!} $$. Multiply these, $$ binom{11}{8} frac{3!}{2!} $$.
$endgroup$
– jayko03
Dec 10 '18 at 3:55
$begingroup$
Your count is correct. However, you should say that you are choosing eight of the eleven positions for the $S$s and $I$s (since the eight letters do not form a block). There is only way to place the $S$s and $I$s in those positions since every $S$ must appear before the first $I$. We must choose one of the remaining three positions for the M. The final two positions must be filled with the two $P$s. We get $$binom{11}{8}binom{3}{1} = binom{11}{8}frac{3!}{1!2!} = binom{11}{8}frac{3!}{2!}$$ Our approaches are equivalent. We just chose to fill the positions in different orders.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 11:45
$begingroup$
In my parenthetical remark in my previous comment, I meant to say that the eight letters may not form a block.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 16:31
add a comment |
$begingroup$
I think this is another approach. I ended up like this. First, group the Ss and Is and treat them as one letter. So there are $$ binom{11}{8} $$ way to organize for Ss and Is. Then there is rest of letters (M, P, P) $$ frac{3!}{2!} $$. Multiply these, $$ binom{11}{8} frac{3!}{2!} $$.
$endgroup$
– jayko03
Dec 10 '18 at 3:55
$begingroup$
Your count is correct. However, you should say that you are choosing eight of the eleven positions for the $S$s and $I$s (since the eight letters do not form a block). There is only way to place the $S$s and $I$s in those positions since every $S$ must appear before the first $I$. We must choose one of the remaining three positions for the M. The final two positions must be filled with the two $P$s. We get $$binom{11}{8}binom{3}{1} = binom{11}{8}frac{3!}{1!2!} = binom{11}{8}frac{3!}{2!}$$ Our approaches are equivalent. We just chose to fill the positions in different orders.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 11:45
$begingroup$
In my parenthetical remark in my previous comment, I meant to say that the eight letters may not form a block.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 16:31
$begingroup$
I think this is another approach. I ended up like this. First, group the Ss and Is and treat them as one letter. So there are $$ binom{11}{8} $$ way to organize for Ss and Is. Then there is rest of letters (M, P, P) $$ frac{3!}{2!} $$. Multiply these, $$ binom{11}{8} frac{3!}{2!} $$.
$endgroup$
– jayko03
Dec 10 '18 at 3:55
$begingroup$
I think this is another approach. I ended up like this. First, group the Ss and Is and treat them as one letter. So there are $$ binom{11}{8} $$ way to organize for Ss and Is. Then there is rest of letters (M, P, P) $$ frac{3!}{2!} $$. Multiply these, $$ binom{11}{8} frac{3!}{2!} $$.
$endgroup$
– jayko03
Dec 10 '18 at 3:55
$begingroup$
Your count is correct. However, you should say that you are choosing eight of the eleven positions for the $S$s and $I$s (since the eight letters do not form a block). There is only way to place the $S$s and $I$s in those positions since every $S$ must appear before the first $I$. We must choose one of the remaining three positions for the M. The final two positions must be filled with the two $P$s. We get $$binom{11}{8}binom{3}{1} = binom{11}{8}frac{3!}{1!2!} = binom{11}{8}frac{3!}{2!}$$ Our approaches are equivalent. We just chose to fill the positions in different orders.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 11:45
$begingroup$
Your count is correct. However, you should say that you are choosing eight of the eleven positions for the $S$s and $I$s (since the eight letters do not form a block). There is only way to place the $S$s and $I$s in those positions since every $S$ must appear before the first $I$. We must choose one of the remaining three positions for the M. The final two positions must be filled with the two $P$s. We get $$binom{11}{8}binom{3}{1} = binom{11}{8}frac{3!}{1!2!} = binom{11}{8}frac{3!}{2!}$$ Our approaches are equivalent. We just chose to fill the positions in different orders.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 11:45
$begingroup$
In my parenthetical remark in my previous comment, I meant to say that the eight letters may not form a block.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 16:31
$begingroup$
In my parenthetical remark in my previous comment, I meant to say that the eight letters may not form a block.
$endgroup$
– N. F. Taussig
Dec 10 '18 at 16:31
add a comment |
$begingroup$
Since all S should come before I consider the word SSSSIIII. Now we need to add the letters MPP. First letter we add can take 9 positikns (anywhere between letters or at ends of SSSSIIII). Second and third can be added at 10 and 11 positions. But P appears twice so we divide by 2!. Final answer is (11!/8!)/2!
answered Dec 13 '18 at 15:51
CuriousCurious
889516
$endgroup$
add a comment |
$begingroup$
Since all S should come before I consider the word SSSSIIII. Now we need to add the letters MPP. First letter we add can take 9 positikns (anywhere between letters or at ends of SSSSIIII). Second and third can be added at 10 and 11 positions. But P appears twice so we divide by 2!. Final answer is (11!/8!)/2!
answered Dec 13 '18 at 15:51
CuriousCurious
889516
$endgroup$
add a comment |
$begingroup$
Since all S should come before I consider the word SSSSIIII. Now we need to add the letters MPP. First letter we add can take 9 positikns (anywhere between letters or at ends of SSSSIIII). Second and third can be added at 10 and 11 positions. But P appears twice so we divide by 2!. Final answer is (11!/8!)/2!
answered Dec 13 '18 at 15:51
CuriousCurious
889516
$endgroup$
Since all S should come before I consider the word SSSSIIII. Now we need to add the letters MPP. First letter we add can take 9 positikns (anywhere between letters or at ends of SSSSIIII). Second and third can be added at 10 and 11 positions. But P appears twice so we divide by 2!. Final answer is (11!/8!)/2!
answered Dec 13 '18 at 15:51
CuriousCurious
889516
answered Dec 13 '18 at 15:51
CuriousCurious
889516
answered Dec 13 '18 at 15:51
CuriousCurious
889516
answered Dec 13 '18 at 15:51
CuriousCurious
889516
889516
add a comment |
add a comment |
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$begingroup$
For restriction you neglected the case like MSSPSPSIIII or SSSMSPIIPII. Think how to accomodate them also.
$endgroup$
– jayant98
Dec 8 '18 at 20:33
$begingroup$
Your idea of considering S and I the same letter is good, but it would yield (11!/8!2!1!) possibilities
$endgroup$
– josinalvo
Dec 8 '18 at 20:40