How to solve a problem about anagrams












-1












$begingroup$



How many anagrams has the word KOMBINATOORIKA with the condition that the $3$ letters O can't be next to each other.




How to get to the answer?



My work. So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got $14! / (3! 2! 2! 2!)$ as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea.






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:54










  • $begingroup$
    Yes there can be two 0s next to each other
    $endgroup$
    – GWL
    Nov 25 '18 at 17:58
















-1












$begingroup$



How many anagrams has the word KOMBINATOORIKA with the condition that the $3$ letters O can't be next to each other.




How to get to the answer?



My work. So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got $14! / (3! 2! 2! 2!)$ as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea.






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:54










  • $begingroup$
    Yes there can be two 0s next to each other
    $endgroup$
    – GWL
    Nov 25 '18 at 17:58














-1












-1








-1





$begingroup$



How many anagrams has the word KOMBINATOORIKA with the condition that the $3$ letters O can't be next to each other.




How to get to the answer?



My work. So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got $14! / (3! 2! 2! 2!)$ as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea.






combinatorics







share|cite|improve this question











$endgroup$





How many anagrams has the word KOMBINATOORIKA with the condition that the $3$ letters O can't be next to each other.




How to get to the answer?



My work. So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got $14! / (3! 2! 2! 2!)$ as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea.






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 18:11









Robert Z

94.6k1063134




94.6k1063134










asked Nov 25 '18 at 17:41









GWLGWL

63




63












  • $begingroup$
    KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:54










  • $begingroup$
    Yes there can be two 0s next to each other
    $endgroup$
    – GWL
    Nov 25 '18 at 17:58


















  • $begingroup$
    KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:54










  • $begingroup$
    Yes there can be two 0s next to each other
    $endgroup$
    – GWL
    Nov 25 '18 at 17:58
















$begingroup$
KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
$endgroup$
– Robert Z
Nov 25 '18 at 17:54




$begingroup$
KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
$endgroup$
– Robert Z
Nov 25 '18 at 17:54












$begingroup$
Yes there can be two 0s next to each other
$endgroup$
– GWL
Nov 25 '18 at 17:58




$begingroup$
Yes there can be two 0s next to each other
$endgroup$
– GWL
Nov 25 '18 at 17:58










2 Answers
2






active

oldest

votes


















0












$begingroup$

Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Use the hint by Robert Z.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 17:53










  • $begingroup$
    So is this correct: $frac{14!}{2^3*3!}-frac{12!}{2^3}=$ 1 756 339 200, Where as what I had before was 137 214 000.
    $endgroup$
    – nafhgood
    Nov 25 '18 at 17:56












  • $begingroup$
    Consider total number of words with O's as a block. So number of letters is 12. Calculate the permutation which will give number of anagrams where the three Os are adjacent. To calculate total number of anagrams use the same method as you have done but include th O's as well. Subtract the former from the latter to get the result.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:04












  • $begingroup$
    It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:06












  • $begingroup$
    I don't understand why my first answer is not right? The reasoning seem correct?
    $endgroup$
    – nafhgood
    Nov 25 '18 at 18:08



















1












$begingroup$

Hint. Consider the total number of anagrams and subtract the number of anagrams where the three Os are adjacent.



P.S. According to your work the total number of anagrams is
$$frac{14!}{3!2!2!2!}$$
which is correct. Now we enumerate the anagrams where we consider the string OOO just as one letter. So the number of letters decreases from $14$ to $12$ and the number of such anagrams is
$$frac{12!}{2!2!2!}.$$
Finally we subtract this result from the first one:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, I figured that much. First I calculated amount of all anagrams (with 3 "o" letters allowed). Then I tried to find amount of anagrams, that contains set of 3 "o"and subtract the answer from total. It was finding the amount of anagrams, that contains set of 3 "o", that caused the problem
    $endgroup$
    – GWL
    Nov 25 '18 at 17:56












  • $begingroup$
    @GWL Please edit your question and share with us what have you found.
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:59










  • $begingroup$
    So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got 14! / 3! 2! 2! 2! as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea
    $endgroup$
    – GWL
    Nov 25 '18 at 18:10











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Use the hint by Robert Z.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 17:53










  • $begingroup$
    So is this correct: $frac{14!}{2^3*3!}-frac{12!}{2^3}=$ 1 756 339 200, Where as what I had before was 137 214 000.
    $endgroup$
    – nafhgood
    Nov 25 '18 at 17:56












  • $begingroup$
    Consider total number of words with O's as a block. So number of letters is 12. Calculate the permutation which will give number of anagrams where the three Os are adjacent. To calculate total number of anagrams use the same method as you have done but include th O's as well. Subtract the former from the latter to get the result.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:04












  • $begingroup$
    It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:06












  • $begingroup$
    I don't understand why my first answer is not right? The reasoning seem correct?
    $endgroup$
    – nafhgood
    Nov 25 '18 at 18:08
















0












$begingroup$

Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Use the hint by Robert Z.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 17:53










  • $begingroup$
    So is this correct: $frac{14!}{2^3*3!}-frac{12!}{2^3}=$ 1 756 339 200, Where as what I had before was 137 214 000.
    $endgroup$
    – nafhgood
    Nov 25 '18 at 17:56












  • $begingroup$
    Consider total number of words with O's as a block. So number of letters is 12. Calculate the permutation which will give number of anagrams where the three Os are adjacent. To calculate total number of anagrams use the same method as you have done but include th O's as well. Subtract the former from the latter to get the result.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:04












  • $begingroup$
    It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:06












  • $begingroup$
    I don't understand why my first answer is not right? The reasoning seem correct?
    $endgroup$
    – nafhgood
    Nov 25 '18 at 18:08














0












0








0





$begingroup$

Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.






share|cite|improve this answer











$endgroup$



Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 '18 at 18:32

























answered Nov 25 '18 at 17:49









nafhgoodnafhgood

1,801422




1,801422








  • 1




    $begingroup$
    Use the hint by Robert Z.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 17:53










  • $begingroup$
    So is this correct: $frac{14!}{2^3*3!}-frac{12!}{2^3}=$ 1 756 339 200, Where as what I had before was 137 214 000.
    $endgroup$
    – nafhgood
    Nov 25 '18 at 17:56












  • $begingroup$
    Consider total number of words with O's as a block. So number of letters is 12. Calculate the permutation which will give number of anagrams where the three Os are adjacent. To calculate total number of anagrams use the same method as you have done but include th O's as well. Subtract the former from the latter to get the result.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:04












  • $begingroup$
    It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:06












  • $begingroup$
    I don't understand why my first answer is not right? The reasoning seem correct?
    $endgroup$
    – nafhgood
    Nov 25 '18 at 18:08














  • 1




    $begingroup$
    Use the hint by Robert Z.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 17:53










  • $begingroup$
    So is this correct: $frac{14!}{2^3*3!}-frac{12!}{2^3}=$ 1 756 339 200, Where as what I had before was 137 214 000.
    $endgroup$
    – nafhgood
    Nov 25 '18 at 17:56












  • $begingroup$
    Consider total number of words with O's as a block. So number of letters is 12. Calculate the permutation which will give number of anagrams where the three Os are adjacent. To calculate total number of anagrams use the same method as you have done but include th O's as well. Subtract the former from the latter to get the result.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:04












  • $begingroup$
    It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 18:06












  • $begingroup$
    I don't understand why my first answer is not right? The reasoning seem correct?
    $endgroup$
    – nafhgood
    Nov 25 '18 at 18:08








1




1




$begingroup$
Use the hint by Robert Z.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 17:53




$begingroup$
Use the hint by Robert Z.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 17:53












$begingroup$
So is this correct: $frac{14!}{2^3*3!}-frac{12!}{2^3}=$ 1 756 339 200, Where as what I had before was 137 214 000.
$endgroup$
– nafhgood
Nov 25 '18 at 17:56






$begingroup$
So is this correct: $frac{14!}{2^3*3!}-frac{12!}{2^3}=$ 1 756 339 200, Where as what I had before was 137 214 000.
$endgroup$
– nafhgood
Nov 25 '18 at 17:56














$begingroup$
Consider total number of words with O's as a block. So number of letters is 12. Calculate the permutation which will give number of anagrams where the three Os are adjacent. To calculate total number of anagrams use the same method as you have done but include th O's as well. Subtract the former from the latter to get the result.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 18:04






$begingroup$
Consider total number of words with O's as a block. So number of letters is 12. Calculate the permutation which will give number of anagrams where the three Os are adjacent. To calculate total number of anagrams use the same method as you have done but include th O's as well. Subtract the former from the latter to get the result.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 18:04














$begingroup$
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
$endgroup$
– Yadati Kiran
Nov 25 '18 at 18:06






$begingroup$
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
$endgroup$
– Yadati Kiran
Nov 25 '18 at 18:06














$begingroup$
I don't understand why my first answer is not right? The reasoning seem correct?
$endgroup$
– nafhgood
Nov 25 '18 at 18:08




$begingroup$
I don't understand why my first answer is not right? The reasoning seem correct?
$endgroup$
– nafhgood
Nov 25 '18 at 18:08











1












$begingroup$

Hint. Consider the total number of anagrams and subtract the number of anagrams where the three Os are adjacent.



P.S. According to your work the total number of anagrams is
$$frac{14!}{3!2!2!2!}$$
which is correct. Now we enumerate the anagrams where we consider the string OOO just as one letter. So the number of letters decreases from $14$ to $12$ and the number of such anagrams is
$$frac{12!}{2!2!2!}.$$
Finally we subtract this result from the first one:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, I figured that much. First I calculated amount of all anagrams (with 3 "o" letters allowed). Then I tried to find amount of anagrams, that contains set of 3 "o"and subtract the answer from total. It was finding the amount of anagrams, that contains set of 3 "o", that caused the problem
    $endgroup$
    – GWL
    Nov 25 '18 at 17:56












  • $begingroup$
    @GWL Please edit your question and share with us what have you found.
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:59










  • $begingroup$
    So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got 14! / 3! 2! 2! 2! as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea
    $endgroup$
    – GWL
    Nov 25 '18 at 18:10
















1












$begingroup$

Hint. Consider the total number of anagrams and subtract the number of anagrams where the three Os are adjacent.



P.S. According to your work the total number of anagrams is
$$frac{14!}{3!2!2!2!}$$
which is correct. Now we enumerate the anagrams where we consider the string OOO just as one letter. So the number of letters decreases from $14$ to $12$ and the number of such anagrams is
$$frac{12!}{2!2!2!}.$$
Finally we subtract this result from the first one:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    yes, I figured that much. First I calculated amount of all anagrams (with 3 "o" letters allowed). Then I tried to find amount of anagrams, that contains set of 3 "o"and subtract the answer from total. It was finding the amount of anagrams, that contains set of 3 "o", that caused the problem
    $endgroup$
    – GWL
    Nov 25 '18 at 17:56












  • $begingroup$
    @GWL Please edit your question and share with us what have you found.
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:59










  • $begingroup$
    So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got 14! / 3! 2! 2! 2! as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea
    $endgroup$
    – GWL
    Nov 25 '18 at 18:10














1












1








1





$begingroup$

Hint. Consider the total number of anagrams and subtract the number of anagrams where the three Os are adjacent.



P.S. According to your work the total number of anagrams is
$$frac{14!}{3!2!2!2!}$$
which is correct. Now we enumerate the anagrams where we consider the string OOO just as one letter. So the number of letters decreases from $14$ to $12$ and the number of such anagrams is
$$frac{12!}{2!2!2!}.$$
Finally we subtract this result from the first one:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$






share|cite|improve this answer











$endgroup$



Hint. Consider the total number of anagrams and subtract the number of anagrams where the three Os are adjacent.



P.S. According to your work the total number of anagrams is
$$frac{14!}{3!2!2!2!}$$
which is correct. Now we enumerate the anagrams where we consider the string OOO just as one letter. So the number of letters decreases from $14$ to $12$ and the number of such anagrams is
$$frac{12!}{2!2!2!}.$$
Finally we subtract this result from the first one:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 '18 at 18:40

























answered Nov 25 '18 at 17:50









Robert ZRobert Z

94.6k1063134




94.6k1063134












  • $begingroup$
    yes, I figured that much. First I calculated amount of all anagrams (with 3 "o" letters allowed). Then I tried to find amount of anagrams, that contains set of 3 "o"and subtract the answer from total. It was finding the amount of anagrams, that contains set of 3 "o", that caused the problem
    $endgroup$
    – GWL
    Nov 25 '18 at 17:56












  • $begingroup$
    @GWL Please edit your question and share with us what have you found.
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:59










  • $begingroup$
    So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got 14! / 3! 2! 2! 2! as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea
    $endgroup$
    – GWL
    Nov 25 '18 at 18:10


















  • $begingroup$
    yes, I figured that much. First I calculated amount of all anagrams (with 3 "o" letters allowed). Then I tried to find amount of anagrams, that contains set of 3 "o"and subtract the answer from total. It was finding the amount of anagrams, that contains set of 3 "o", that caused the problem
    $endgroup$
    – GWL
    Nov 25 '18 at 17:56












  • $begingroup$
    @GWL Please edit your question and share with us what have you found.
    $endgroup$
    – Robert Z
    Nov 25 '18 at 17:59










  • $begingroup$
    So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got 14! / 3! 2! 2! 2! as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea
    $endgroup$
    – GWL
    Nov 25 '18 at 18:10
















$begingroup$
yes, I figured that much. First I calculated amount of all anagrams (with 3 "o" letters allowed). Then I tried to find amount of anagrams, that contains set of 3 "o"and subtract the answer from total. It was finding the amount of anagrams, that contains set of 3 "o", that caused the problem
$endgroup$
– GWL
Nov 25 '18 at 17:56






$begingroup$
yes, I figured that much. First I calculated amount of all anagrams (with 3 "o" letters allowed). Then I tried to find amount of anagrams, that contains set of 3 "o"and subtract the answer from total. It was finding the amount of anagrams, that contains set of 3 "o", that caused the problem
$endgroup$
– GWL
Nov 25 '18 at 17:56














$begingroup$
@GWL Please edit your question and share with us what have you found.
$endgroup$
– Robert Z
Nov 25 '18 at 17:59




$begingroup$
@GWL Please edit your question and share with us what have you found.
$endgroup$
– Robert Z
Nov 25 '18 at 17:59












$begingroup$
So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got 14! / 3! 2! 2! 2! as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea
$endgroup$
– GWL
Nov 25 '18 at 18:10




$begingroup$
So the first thing I did, was to find how many anagrams the word "KOMBINATOORIKA" has in total, with no extra conditions I got 14! / 3! 2! 2! 2! as an answer. Now I try to find how many different anagram contain combination "OOO". I could subtract answer from my first answer. That would give me end result. This is the idea
$endgroup$
– GWL
Nov 25 '18 at 18:10


















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