How to solve a problem about anagrams
up vote
-1
down vote
favorite
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
edited Nov 25 at 18:11
Robert Z
91.9k1058129
asked Nov 25 at 17:41
GWL
63
add a comment |
up vote
-1
down vote
favorite
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
edited Nov 25 at 18:11
Robert Z
91.9k1058129
asked Nov 25 at 17:41
GWL
63
KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
– Robert Z
Nov 25 at 17:54
Yes there can be two 0s next to each other
– GWL
Nov 25 at 17:58
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
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
edited Nov 25 at 18:11
Robert Z
91.9k1058129
asked Nov 25 at 17:41
GWL
63
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
edited Nov 25 at 18:11
Robert Z
91.9k1058129
asked Nov 25 at 17:41
GWL
63
edited Nov 25 at 18:11
Robert Z
91.9k1058129
asked Nov 25 at 17:41
GWL
63
edited Nov 25 at 18:11
Robert Z
91.9k1058129
edited Nov 25 at 18:11
Robert Z
91.9k1058129
edited Nov 25 at 18:11
Robert Z
91.9k1058129
91.9k1058129
asked Nov 25 at 17:41
GWL
63
asked Nov 25 at 17:41
GWL
63
asked Nov 25 at 17:41
GWL
63
63
KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
– Robert Z
Nov 25 at 17:54
Yes there can be two 0s next to each other
– GWL
Nov 25 at 17:58
add a comment |
KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
– Robert Z
Nov 25 at 17:54
Yes there can be two 0s next to each other
– GWL
Nov 25 at 17:58
KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
– Robert Z
Nov 25 at 17:54
KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
– Robert Z
Nov 25 at 17:54
Yes there can be two 0s next to each other
– GWL
Nov 25 at 17:58
Yes there can be two 0s next to each other
– GWL
Nov 25 at 17:58
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.
answered Nov 25 at 17:49
mathnoob
1,563321
1
Use the hint by Robert Z.
– Yadati Kiran
Nov 25 at 17:53
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.
– mathnoob
Nov 25 at 17:56
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.
– Yadati Kiran
Nov 25 at 18:04
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
– Yadati Kiran
Nov 25 at 18:06
I don't understand why my first answer is not right? The reasoning seem correct?
– mathnoob
Nov 25 at 18:08
|
show 4 more comments
up vote
1
down vote
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:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$
answered Nov 25 at 17:50
Robert Z
91.9k1058129
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
– GWL
Nov 25 at 17:56
@GWL Please edit your question and share with us what have you found.
– Robert Z
Nov 25 at 17:59
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
– GWL
Nov 25 at 18:10
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.
answered Nov 25 at 17:49
mathnoob
1,563321
1
Use the hint by Robert Z.
– Yadati Kiran
Nov 25 at 17:53
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.
– mathnoob
Nov 25 at 17:56
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.
– Yadati Kiran
Nov 25 at 18:04
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
– Yadati Kiran
Nov 25 at 18:06
I don't understand why my first answer is not right? The reasoning seem correct?
– mathnoob
Nov 25 at 18:08
|
show 4 more comments
up vote
0
down vote
accepted
Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.
answered Nov 25 at 17:49
mathnoob
1,563321
1
Use the hint by Robert Z.
– Yadati Kiran
Nov 25 at 17:53
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.
– mathnoob
Nov 25 at 17:56
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.
– Yadati Kiran
Nov 25 at 18:04
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
– Yadati Kiran
Nov 25 at 18:06
I don't understand why my first answer is not right? The reasoning seem correct?
– mathnoob
Nov 25 at 18:08
|
show 4 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.
answered Nov 25 at 17:49
mathnoob
1,563321
Ok the solution is $frac{14!}{2^3*6}-frac{12!}{2^3}$.
answered Nov 25 at 17:49
mathnoob
1,563321
edited Nov 25 at 18:32
answered Nov 25 at 17:49
mathnoob
1,563321
answered Nov 25 at 17:49
mathnoob
1,563321
answered Nov 25 at 17:49
mathnoob
1,563321
1,563321
1
Use the hint by Robert Z.
– Yadati Kiran
Nov 25 at 17:53
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.
– mathnoob
Nov 25 at 17:56
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.
– Yadati Kiran
Nov 25 at 18:04
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
– Yadati Kiran
Nov 25 at 18:06
I don't understand why my first answer is not right? The reasoning seem correct?
– mathnoob
Nov 25 at 18:08
|
show 4 more comments
1
Use the hint by Robert Z.
– Yadati Kiran
Nov 25 at 17:53
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.
– mathnoob
Nov 25 at 17:56
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.
– Yadati Kiran
Nov 25 at 18:04
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
– Yadati Kiran
Nov 25 at 18:06
I don't understand why my first answer is not right? The reasoning seem correct?
– mathnoob
Nov 25 at 18:08
1
1
Use the hint by Robert Z.
– Yadati Kiran
Nov 25 at 17:53
Use the hint by Robert Z.
– Yadati Kiran
Nov 25 at 17:53
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.
– mathnoob
Nov 25 at 17:56
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.
– mathnoob
Nov 25 at 17:56
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.
– Yadati Kiran
Nov 25 at 18:04
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.
– Yadati Kiran
Nov 25 at 18:04
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
– Yadati Kiran
Nov 25 at 18:06
It seems you have got it right. $dfrac{14!}{2^3cdot 3!}-dfrac{12!}{2^3}=$
– Yadati Kiran
Nov 25 at 18:06
I don't understand why my first answer is not right? The reasoning seem correct?
– mathnoob
Nov 25 at 18:08
I don't understand why my first answer is not right? The reasoning seem correct?
– mathnoob
Nov 25 at 18:08
|
show 4 more comments
up vote
1
down vote
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:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$
answered Nov 25 at 17:50
Robert Z
91.9k1058129
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
– GWL
Nov 25 at 17:56
@GWL Please edit your question and share with us what have you found.
– Robert Z
Nov 25 at 17:59
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
– GWL
Nov 25 at 18:10
add a comment |
up vote
1
down vote
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:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$
answered Nov 25 at 17:50
Robert Z
91.9k1058129
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
– GWL
Nov 25 at 17:56
@GWL Please edit your question and share with us what have you found.
– Robert Z
Nov 25 at 17:59
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
– GWL
Nov 25 at 18:10
add a comment |
up vote
1
down vote
up vote
1
down vote
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:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$
answered Nov 25 at 17:50
Robert Z
91.9k1058129
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:
$$frac{14!}{3!2!2!2!}-frac{12!}{2!2!2!}.$$
answered Nov 25 at 17:50
Robert Z
91.9k1058129
edited Nov 25 at 18:40
answered Nov 25 at 17:50
Robert Z
91.9k1058129
answered Nov 25 at 17:50
Robert Z
91.9k1058129
answered Nov 25 at 17:50
Robert Z
91.9k1058129
91.9k1058129
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
– GWL
Nov 25 at 17:56
@GWL Please edit your question and share with us what have you found.
– Robert Z
Nov 25 at 17:59
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
– GWL
Nov 25 at 18:10
add a comment |
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
– GWL
Nov 25 at 17:56
@GWL Please edit your question and share with us what have you found.
– Robert Z
Nov 25 at 17:59
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
– GWL
Nov 25 at 18:10
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
– GWL
Nov 25 at 17:56
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
– GWL
Nov 25 at 17:56
@GWL Please edit your question and share with us what have you found.
– Robert Z
Nov 25 at 17:59
@GWL Please edit your question and share with us what have you found.
– Robert Z
Nov 25 at 17:59
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
– GWL
Nov 25 at 18:10
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
– GWL
Nov 25 at 18:10
add a comment |
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KOOOMBINATRIKA is not admitted. What about KOMBINATOORIKA? Can two Os be next to each other?
– Robert Z
Nov 25 at 17:54
Yes there can be two 0s next to each other
– GWL
Nov 25 at 17:58