If all permutations of word âFATIMAHâ are written in lexicographic order. What would be the 1444th word?
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Following is my question:
If all permutations of word "FATIMAH" are written in lexicographic order. What would be the 1444th word you write?
Here is my solution:
Firstly arrange them in alphabetic order: $A,F,H,I,M,T$
Now all the permutations when first letter is A $= 6! = 720$
All the permutations when first letter is F $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is H $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is I $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is M $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is T $= dfrac6!2! = 360$ (since
A is repeated twice)
Total words up to H $= 1440$
$1441$st word = IAAFHMT
$1442$nd word = IAAFHTM
$1443$rd word = IAAFTHM
$1444$th word = IAATFHM
Please tell whether this is correct or not.
combinatorics permutations
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up vote
3
down vote
favorite
Following is my question:
If all permutations of word "FATIMAH" are written in lexicographic order. What would be the 1444th word you write?
Here is my solution:
Firstly arrange them in alphabetic order: $A,F,H,I,M,T$
Now all the permutations when first letter is A $= 6! = 720$
All the permutations when first letter is F $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is H $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is I $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is M $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is T $= dfrac6!2! = 360$ (since
A is repeated twice)
Total words up to H $= 1440$
$1441$st word = IAAFHMT
$1442$nd word = IAAFHTM
$1443$rd word = IAAFTHM
$1444$th word = IAATFHM
Please tell whether this is correct or not.
combinatorics permutations
Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
â N. F. Taussig
Sep 13 at 11:15
1
I think 1443 will be $IAAFMHT$ and 1444 will be $IAAFMTH$
â Vasya
Sep 13 at 11:20
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up vote
3
down vote
favorite
up vote
3
down vote
favorite
Following is my question:
If all permutations of word "FATIMAH" are written in lexicographic order. What would be the 1444th word you write?
Here is my solution:
Firstly arrange them in alphabetic order: $A,F,H,I,M,T$
Now all the permutations when first letter is A $= 6! = 720$
All the permutations when first letter is F $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is H $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is I $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is M $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is T $= dfrac6!2! = 360$ (since
A is repeated twice)
Total words up to H $= 1440$
$1441$st word = IAAFHMT
$1442$nd word = IAAFHTM
$1443$rd word = IAAFTHM
$1444$th word = IAATFHM
Please tell whether this is correct or not.
combinatorics permutations
Following is my question:
If all permutations of word "FATIMAH" are written in lexicographic order. What would be the 1444th word you write?
Here is my solution:
Firstly arrange them in alphabetic order: $A,F,H,I,M,T$
Now all the permutations when first letter is A $= 6! = 720$
All the permutations when first letter is F $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is H $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is I $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is M $= dfrac6!2! = 360$ (since A is repeated twice)
All the permutations when first letter is T $= dfrac6!2! = 360$ (since
A is repeated twice)
Total words up to H $= 1440$
$1441$st word = IAAFHMT
$1442$nd word = IAAFHTM
$1443$rd word = IAAFTHM
$1444$th word = IAATFHM
Please tell whether this is correct or not.
combinatorics permutations
combinatorics permutations
edited Sep 13 at 11:14
N. F. Taussig
40.4k93253
40.4k93253
asked Sep 13 at 11:06
puffles
554
554
Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
â N. F. Taussig
Sep 13 at 11:15
1
I think 1443 will be $IAAFMHT$ and 1444 will be $IAAFMTH$
â Vasya
Sep 13 at 11:20
add a comment |Â
Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
â N. F. Taussig
Sep 13 at 11:15
1
I think 1443 will be $IAAFMHT$ and 1444 will be $IAAFMTH$
â Vasya
Sep 13 at 11:20
Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
â N. F. Taussig
Sep 13 at 11:15
Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
â N. F. Taussig
Sep 13 at 11:15
1
1
I think 1443 will be $IAAFMHT$ and 1444 will be $IAAFMTH$
â Vasya
Sep 13 at 11:20
I think 1443 will be $IAAFMHT$ and 1444 will be $IAAFMTH$
â Vasya
Sep 13 at 11:20
add a comment |Â
3 Answers
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Your answer is correct until the 1443rd word.
You can think of it this way:
We know that the 1443rd word will be (obviously) $IAA****$. Now we can label $F=1, H=2, M=3, T=4$
And now the question simplifies into arranging 4-digit numbers made by $1,2,3,4$ in ascending order and taking the 4th one.
1: $1234$
2: $1243$
3: $1324$
4: $1342$
Now we can change the numbers back to the corresponding letters, and we will get $IAAFMTH$. $ _square$
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up vote
2
down vote
There are $1440$ words with first letter $A$, $F$ or $H$, as you've correctly found out.
Using the same tactic you used to count words beginning with a single letter, count the number of words beginning with $IAAF$. You will see that there are actually six. You only wrote down three of them before moving on to $IAAT$ (which is also wrong, because $IAAH$ and $IAAM$ both come before $IAAT$ lexicographically).
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Almost. Your argument is correct until you reach the $1443$rd word. Since F precedes H, H precedes M, and M precedes T in the alphabet, All the words beginning with IAAF must precede the first word beginning with IAAH, all the words beginning with IAAH must precede the first word beginning with IAAM, and all the words beginning with IAAM must precede the first word beginning with IAAT. The words beginning with IAAF are, in lexicographical order, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH, IAAFTHM, IAAAFTMH. Thus, the $1441$st word, $1442$nd word, $1443$rd word, and $1444$th word are, respectively, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH.
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your answer is correct until the 1443rd word.
You can think of it this way:
We know that the 1443rd word will be (obviously) $IAA****$. Now we can label $F=1, H=2, M=3, T=4$
And now the question simplifies into arranging 4-digit numbers made by $1,2,3,4$ in ascending order and taking the 4th one.
1: $1234$
2: $1243$
3: $1324$
4: $1342$
Now we can change the numbers back to the corresponding letters, and we will get $IAAFMTH$. $ _square$
add a comment |Â
up vote
2
down vote
accepted
Your answer is correct until the 1443rd word.
You can think of it this way:
We know that the 1443rd word will be (obviously) $IAA****$. Now we can label $F=1, H=2, M=3, T=4$
And now the question simplifies into arranging 4-digit numbers made by $1,2,3,4$ in ascending order and taking the 4th one.
1: $1234$
2: $1243$
3: $1324$
4: $1342$
Now we can change the numbers back to the corresponding letters, and we will get $IAAFMTH$. $ _square$
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your answer is correct until the 1443rd word.
You can think of it this way:
We know that the 1443rd word will be (obviously) $IAA****$. Now we can label $F=1, H=2, M=3, T=4$
And now the question simplifies into arranging 4-digit numbers made by $1,2,3,4$ in ascending order and taking the 4th one.
1: $1234$
2: $1243$
3: $1324$
4: $1342$
Now we can change the numbers back to the corresponding letters, and we will get $IAAFMTH$. $ _square$
Your answer is correct until the 1443rd word.
You can think of it this way:
We know that the 1443rd word will be (obviously) $IAA****$. Now we can label $F=1, H=2, M=3, T=4$
And now the question simplifies into arranging 4-digit numbers made by $1,2,3,4$ in ascending order and taking the 4th one.
1: $1234$
2: $1243$
3: $1324$
4: $1342$
Now we can change the numbers back to the corresponding letters, and we will get $IAAFMTH$. $ _square$
answered Sep 13 at 11:54
Vee Hua Zhi
76419
76419
add a comment |Â
add a comment |Â
up vote
2
down vote
There are $1440$ words with first letter $A$, $F$ or $H$, as you've correctly found out.
Using the same tactic you used to count words beginning with a single letter, count the number of words beginning with $IAAF$. You will see that there are actually six. You only wrote down three of them before moving on to $IAAT$ (which is also wrong, because $IAAH$ and $IAAM$ both come before $IAAT$ lexicographically).
add a comment |Â
up vote
2
down vote
There are $1440$ words with first letter $A$, $F$ or $H$, as you've correctly found out.
Using the same tactic you used to count words beginning with a single letter, count the number of words beginning with $IAAF$. You will see that there are actually six. You only wrote down three of them before moving on to $IAAT$ (which is also wrong, because $IAAH$ and $IAAM$ both come before $IAAT$ lexicographically).
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There are $1440$ words with first letter $A$, $F$ or $H$, as you've correctly found out.
Using the same tactic you used to count words beginning with a single letter, count the number of words beginning with $IAAF$. You will see that there are actually six. You only wrote down three of them before moving on to $IAAT$ (which is also wrong, because $IAAH$ and $IAAM$ both come before $IAAT$ lexicographically).
There are $1440$ words with first letter $A$, $F$ or $H$, as you've correctly found out.
Using the same tactic you used to count words beginning with a single letter, count the number of words beginning with $IAAF$. You will see that there are actually six. You only wrote down three of them before moving on to $IAAT$ (which is also wrong, because $IAAH$ and $IAAM$ both come before $IAAT$ lexicographically).
answered Sep 13 at 11:20
Arthur
103k797179
103k797179
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1
down vote
Almost. Your argument is correct until you reach the $1443$rd word. Since F precedes H, H precedes M, and M precedes T in the alphabet, All the words beginning with IAAF must precede the first word beginning with IAAH, all the words beginning with IAAH must precede the first word beginning with IAAM, and all the words beginning with IAAM must precede the first word beginning with IAAT. The words beginning with IAAF are, in lexicographical order, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH, IAAFTHM, IAAAFTMH. Thus, the $1441$st word, $1442$nd word, $1443$rd word, and $1444$th word are, respectively, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH.
add a comment |Â
up vote
1
down vote
Almost. Your argument is correct until you reach the $1443$rd word. Since F precedes H, H precedes M, and M precedes T in the alphabet, All the words beginning with IAAF must precede the first word beginning with IAAH, all the words beginning with IAAH must precede the first word beginning with IAAM, and all the words beginning with IAAM must precede the first word beginning with IAAT. The words beginning with IAAF are, in lexicographical order, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH, IAAFTHM, IAAAFTMH. Thus, the $1441$st word, $1442$nd word, $1443$rd word, and $1444$th word are, respectively, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Almost. Your argument is correct until you reach the $1443$rd word. Since F precedes H, H precedes M, and M precedes T in the alphabet, All the words beginning with IAAF must precede the first word beginning with IAAH, all the words beginning with IAAH must precede the first word beginning with IAAM, and all the words beginning with IAAM must precede the first word beginning with IAAT. The words beginning with IAAF are, in lexicographical order, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH, IAAFTHM, IAAAFTMH. Thus, the $1441$st word, $1442$nd word, $1443$rd word, and $1444$th word are, respectively, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH.
Almost. Your argument is correct until you reach the $1443$rd word. Since F precedes H, H precedes M, and M precedes T in the alphabet, All the words beginning with IAAF must precede the first word beginning with IAAH, all the words beginning with IAAH must precede the first word beginning with IAAM, and all the words beginning with IAAM must precede the first word beginning with IAAT. The words beginning with IAAF are, in lexicographical order, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH, IAAFTHM, IAAAFTMH. Thus, the $1441$st word, $1442$nd word, $1443$rd word, and $1444$th word are, respectively, IAAFHMT, IAAFHTM, IAAFMHT, IAAFMTH.
edited Sep 13 at 11:36
answered Sep 13 at 11:20
N. F. Taussig
40.4k93253
40.4k93253
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Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
â N. F. Taussig
Sep 13 at 11:15
1
I think 1443 will be $IAAFMHT$ and 1444 will be $IAAFMTH$
â Vasya
Sep 13 at 11:20