In how many ways can the letters of the word ALGEBRA be rearranged?












0












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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.










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  • $begingroup$
    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


















0












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 17:00







Rajdeep Biswas

















asked Dec 15 '18 at 16:51









Rajdeep BiswasRajdeep Biswas

284




284












  • $begingroup$
    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






  • 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




$begingroup$
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












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