In how many ways can the letters of the word ALGEBRA be rearranged?
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In how many ways, can the letters of the word ALGEBRA be rearranged with condition: the two As never appear next to each other.
The way I think of a solution is: total number of permutations of the word ALGEBRA (accounting for the two As) - the total number of permutations of the word ALGEBRA where the two As do occur together.
The answer stood out to be 1800. Can someone confirm this?
EDIT: elaborated solution -
1. total number of permutations of the word ALGEBRA with the two As accounted for: 7! / 2! = 2520
2. total number of permutations of the word ALGEBRA with the two As always occurring together("AA",L,G,E,B,R being the distinct elements): 6! = 720
hence answer is 1 minus 2 i.e 1800.
combinatorics permutations combinations
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add a comment |
$begingroup$
In how many ways, can the letters of the word ALGEBRA be rearranged with condition: the two As never appear next to each other.
The way I think of a solution is: total number of permutations of the word ALGEBRA (accounting for the two As) - the total number of permutations of the word ALGEBRA where the two As do occur together.
The answer stood out to be 1800. Can someone confirm this?
EDIT: elaborated solution -
1. total number of permutations of the word ALGEBRA with the two As accounted for: 7! / 2! = 2520
2. total number of permutations of the word ALGEBRA with the two As always occurring together("AA",L,G,E,B,R being the distinct elements): 6! = 720
hence answer is 1 minus 2 i.e 1800.
combinatorics permutations combinations
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Welcome to stackexchange. Please edit the question to show us just how you came up with that answer.
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– Ethan Bolker
Dec 15 '18 at 16:54
2
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Looks OK to me.
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– Ethan Bolker
Dec 15 '18 at 17:03
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This site is full of similar questions, e.g., here and the related questions. It might be useful to compare with them to obtain confirmation yourself (you don't need confirmation by others, you want to see it yourself).
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– Dietrich Burde
Dec 15 '18 at 17:23
add a comment |
$begingroup$
In how many ways, can the letters of the word ALGEBRA be rearranged with condition: the two As never appear next to each other.
The way I think of a solution is: total number of permutations of the word ALGEBRA (accounting for the two As) - the total number of permutations of the word ALGEBRA where the two As do occur together.
The answer stood out to be 1800. Can someone confirm this?
EDIT: elaborated solution -
1. total number of permutations of the word ALGEBRA with the two As accounted for: 7! / 2! = 2520
2. total number of permutations of the word ALGEBRA with the two As always occurring together("AA",L,G,E,B,R being the distinct elements): 6! = 720
hence answer is 1 minus 2 i.e 1800.
combinatorics permutations combinations
$endgroup$
In how many ways, can the letters of the word ALGEBRA be rearranged with condition: the two As never appear next to each other.
The way I think of a solution is: total number of permutations of the word ALGEBRA (accounting for the two As) - the total number of permutations of the word ALGEBRA where the two As do occur together.
The answer stood out to be 1800. Can someone confirm this?
EDIT: elaborated solution -
1. total number of permutations of the word ALGEBRA with the two As accounted for: 7! / 2! = 2520
2. total number of permutations of the word ALGEBRA with the two As always occurring together("AA",L,G,E,B,R being the distinct elements): 6! = 720
hence answer is 1 minus 2 i.e 1800.
combinatorics permutations combinations
combinatorics permutations combinations
edited Dec 15 '18 at 17:00
Rajdeep Biswas
asked Dec 15 '18 at 16:51
Rajdeep BiswasRajdeep Biswas
284
284
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Welcome to stackexchange. Please edit the question to show us just how you came up with that answer.
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– Ethan Bolker
Dec 15 '18 at 16:54
2
$begingroup$
Looks OK to me.
$endgroup$
– Ethan Bolker
Dec 15 '18 at 17:03
$begingroup$
This site is full of similar questions, e.g., here and the related questions. It might be useful to compare with them to obtain confirmation yourself (you don't need confirmation by others, you want to see it yourself).
$endgroup$
– Dietrich Burde
Dec 15 '18 at 17:23
add a comment |
$begingroup$
Welcome to stackexchange. Please edit the question to show us just how you came up with that answer.
$endgroup$
– Ethan Bolker
Dec 15 '18 at 16:54
2
$begingroup$
Looks OK to me.
$endgroup$
– Ethan Bolker
Dec 15 '18 at 17:03
$begingroup$
This site is full of similar questions, e.g., here and the related questions. It might be useful to compare with them to obtain confirmation yourself (you don't need confirmation by others, you want to see it yourself).
$endgroup$
– Dietrich Burde
Dec 15 '18 at 17:23
$begingroup$
Welcome to stackexchange. Please edit the question to show us just how you came up with that answer.
$endgroup$
– Ethan Bolker
Dec 15 '18 at 16:54
$begingroup$
Welcome to stackexchange. Please edit the question to show us just how you came up with that answer.
$endgroup$
– Ethan Bolker
Dec 15 '18 at 16:54
2
2
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Looks OK to me.
$endgroup$
– Ethan Bolker
Dec 15 '18 at 17:03
$begingroup$
Looks OK to me.
$endgroup$
– Ethan Bolker
Dec 15 '18 at 17:03
$begingroup$
This site is full of similar questions, e.g., here and the related questions. It might be useful to compare with them to obtain confirmation yourself (you don't need confirmation by others, you want to see it yourself).
$endgroup$
– Dietrich Burde
Dec 15 '18 at 17:23
$begingroup$
This site is full of similar questions, e.g., here and the related questions. It might be useful to compare with them to obtain confirmation yourself (you don't need confirmation by others, you want to see it yourself).
$endgroup$
– Dietrich Burde
Dec 15 '18 at 17:23
add a comment |
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$begingroup$
Welcome to stackexchange. Please edit the question to show us just how you came up with that answer.
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– Ethan Bolker
Dec 15 '18 at 16:54
2
$begingroup$
Looks OK to me.
$endgroup$
– Ethan Bolker
Dec 15 '18 at 17:03
$begingroup$
This site is full of similar questions, e.g., here and the related questions. It might be useful to compare with them to obtain confirmation yourself (you don't need confirmation by others, you want to see it yourself).
$endgroup$
– Dietrich Burde
Dec 15 '18 at 17:23