Number of permutations of the word “ABCDEFGHI”

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up vote
2
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How many permutations of the letters ABCDEFGHI are there...



  1. That end with any letter other than C.

  2. That contain the string HI

  3. That contain the string ACD

  4. That contain the strings AB, DE and GH

  5. If letter A is somewhere to the left of letter E

  6. If letter A is somewhere to the left of letter E and there is exactly one letter between A and E

Question 1



Total number of permutations - Permutations where the letter ends in C:



9! - 8! = 322560



Question 2



Treat "HI" as singular letter and calculate permutation as usual:



8! = 40320



Question 3



Treat "ACD" as singular letter:



7! = 5040



Question 4



Treat "AB", "DE", "GH" as singular letter:



6! = 720



Question 5, 6



This is where I hit a wall. How do I know the position of A in relation to B? I feel that I won't understand the answer even if I see it.



Is my answer to Q1 - Q4 correct? What is the key to solving Q5 and Q6?






combinatorics permutations







share|cite|improve this question





















  • are you familiar with the "stars and bars" method?
    – RGS
    4 hours ago










  • @RGS I saw a tutorial once, but I need to refresh my memory.
    – potatoguy
    4 hours ago














up vote
2
down vote

favorite












How many permutations of the letters ABCDEFGHI are there...



  1. That end with any letter other than C.

  2. That contain the string HI

  3. That contain the string ACD

  4. That contain the strings AB, DE and GH

  5. If letter A is somewhere to the left of letter E

  6. If letter A is somewhere to the left of letter E and there is exactly one letter between A and E

Question 1



Total number of permutations - Permutations where the letter ends in C:



9! - 8! = 322560



Question 2



Treat "HI" as singular letter and calculate permutation as usual:



8! = 40320



Question 3



Treat "ACD" as singular letter:



7! = 5040



Question 4



Treat "AB", "DE", "GH" as singular letter:



6! = 720



Question 5, 6



This is where I hit a wall. How do I know the position of A in relation to B? I feel that I won't understand the answer even if I see it.



Is my answer to Q1 - Q4 correct? What is the key to solving Q5 and Q6?






combinatorics permutations







share|cite|improve this question





















  • are you familiar with the "stars and bars" method?
    – RGS
    4 hours ago










  • @RGS I saw a tutorial once, but I need to refresh my memory.
    – potatoguy
    4 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











How many permutations of the letters ABCDEFGHI are there...



  1. That end with any letter other than C.

  2. That contain the string HI

  3. That contain the string ACD

  4. That contain the strings AB, DE and GH

  5. If letter A is somewhere to the left of letter E

  6. If letter A is somewhere to the left of letter E and there is exactly one letter between A and E

Question 1



Total number of permutations - Permutations where the letter ends in C:



9! - 8! = 322560



Question 2



Treat "HI" as singular letter and calculate permutation as usual:



8! = 40320



Question 3



Treat "ACD" as singular letter:



7! = 5040



Question 4



Treat "AB", "DE", "GH" as singular letter:



6! = 720



Question 5, 6



This is where I hit a wall. How do I know the position of A in relation to B? I feel that I won't understand the answer even if I see it.



Is my answer to Q1 - Q4 correct? What is the key to solving Q5 and Q6?






combinatorics permutations







share|cite|improve this question













How many permutations of the letters ABCDEFGHI are there...



  1. That end with any letter other than C.

  2. That contain the string HI

  3. That contain the string ACD

  4. That contain the strings AB, DE and GH

  5. If letter A is somewhere to the left of letter E

  6. If letter A is somewhere to the left of letter E and there is exactly one letter between A and E

Question 1



Total number of permutations - Permutations where the letter ends in C:



9! - 8! = 322560



Question 2



Treat "HI" as singular letter and calculate permutation as usual:



8! = 40320



Question 3



Treat "ACD" as singular letter:



7! = 5040



Question 4



Treat "AB", "DE", "GH" as singular letter:



6! = 720



Question 5, 6



This is where I hit a wall. How do I know the position of A in relation to B? I feel that I won't understand the answer even if I see it.



Is my answer to Q1 - Q4 correct? What is the key to solving Q5 and Q6?






combinatorics permutations




combinatorics permutations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









potatoguy

212




212











  • are you familiar with the "stars and bars" method?
    – RGS
    4 hours ago










  • @RGS I saw a tutorial once, but I need to refresh my memory.
    – potatoguy
    4 hours ago
















  • are you familiar with the "stars and bars" method?
    – RGS
    4 hours ago










  • @RGS I saw a tutorial once, but I need to refresh my memory.
    – potatoguy
    4 hours ago















are you familiar with the "stars and bars" method?
– RGS
4 hours ago




are you familiar with the "stars and bars" method?
– RGS
4 hours ago












@RGS I saw a tutorial once, but I need to refresh my memory.
– potatoguy
4 hours ago




@RGS I saw a tutorial once, but I need to refresh my memory.
– potatoguy
4 hours ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote













I find your answers to Q1 through Q4 fine!



To answer Q5, start by writing down the A and the E:



$$_ A _ E _$$



where the underscores represent the three boxes in which you can fit the other letters. Can you count in how many ways those three boxes can be filled?



To answer Q6, you go for a similar reasoning. Write $A$ and $E$:



$$_ A - E _$$



but now you start by assigning a single letter to $-$. Then you distribute all the other letters among the two boxes $_$.






share|cite|improve this answer




















  • I see! So given n amount of letters besides A and E, I need to find the number of permutations of how they can be placed inside these three spaces. I'm guessing it should be something like C(7+3-1, 2)?
    – potatoguy
    4 hours ago











  • Yes but then be careful, as you also need to permutate them to fix their order! Using the combinations with C you only decide which letters go to which box and you don't get to count their orderings. Give it a thought. If you can't make it I will further develop my answer.
    – RGS
    4 hours ago

















up vote
2
down vote













1-4 look fine.



5)



In exactly $frac 12$ of all permuations will A be to the left of E, and the other half it will be to the right.



$frac 9!2$



6) We have a sequence $AxE$ there are $7$ values that $x$ can be. And then think of $AxE$ as a single letter.



$7cdot 7!$






share|cite|improve this answer




















  • Your answer is quite clever. I think it would've been better, however, if you made the OP get there instead of just showing the solution :'(
    – RGS
    4 hours ago










  • Thanks for writing the explanation as well. It's important for me to know why the solution is correct. In hindsight, I really overcomplicated it in my head.
    – potatoguy
    4 hours ago






  • 1




    @RGS Thanks for having so much faith in me. It still took me some time to understand Doug's answer.
    – potatoguy
    4 hours ago










Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote













I find your answers to Q1 through Q4 fine!



To answer Q5, start by writing down the A and the E:



$$_ A _ E _$$



where the underscores represent the three boxes in which you can fit the other letters. Can you count in how many ways those three boxes can be filled?



To answer Q6, you go for a similar reasoning. Write $A$ and $E$:



$$_ A - E _$$



but now you start by assigning a single letter to $-$. Then you distribute all the other letters among the two boxes $_$.






share|cite|improve this answer




















  • I see! So given n amount of letters besides A and E, I need to find the number of permutations of how they can be placed inside these three spaces. I'm guessing it should be something like C(7+3-1, 2)?
    – potatoguy
    4 hours ago











  • Yes but then be careful, as you also need to permutate them to fix their order! Using the combinations with C you only decide which letters go to which box and you don't get to count their orderings. Give it a thought. If you can't make it I will further develop my answer.
    – RGS
    4 hours ago














up vote
3
down vote













I find your answers to Q1 through Q4 fine!



To answer Q5, start by writing down the A and the E:



$$_ A _ E _$$



where the underscores represent the three boxes in which you can fit the other letters. Can you count in how many ways those three boxes can be filled?



To answer Q6, you go for a similar reasoning. Write $A$ and $E$:



$$_ A - E _$$



but now you start by assigning a single letter to $-$. Then you distribute all the other letters among the two boxes $_$.






share|cite|improve this answer




















  • I see! So given n amount of letters besides A and E, I need to find the number of permutations of how they can be placed inside these three spaces. I'm guessing it should be something like C(7+3-1, 2)?
    – potatoguy
    4 hours ago











  • Yes but then be careful, as you also need to permutate them to fix their order! Using the combinations with C you only decide which letters go to which box and you don't get to count their orderings. Give it a thought. If you can't make it I will further develop my answer.
    – RGS
    4 hours ago












up vote
3
down vote










up vote
3
down vote









I find your answers to Q1 through Q4 fine!



To answer Q5, start by writing down the A and the E:



$$_ A _ E _$$



where the underscores represent the three boxes in which you can fit the other letters. Can you count in how many ways those three boxes can be filled?



To answer Q6, you go for a similar reasoning. Write $A$ and $E$:



$$_ A - E _$$



but now you start by assigning a single letter to $-$. Then you distribute all the other letters among the two boxes $_$.






share|cite|improve this answer












I find your answers to Q1 through Q4 fine!



To answer Q5, start by writing down the A and the E:



$$_ A _ E _$$



where the underscores represent the three boxes in which you can fit the other letters. Can you count in how many ways those three boxes can be filled?



To answer Q6, you go for a similar reasoning. Write $A$ and $E$:



$$_ A - E _$$



but now you start by assigning a single letter to $-$. Then you distribute all the other letters among the two boxes $_$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









RGS

8,59211129




8,59211129











  • I see! So given n amount of letters besides A and E, I need to find the number of permutations of how they can be placed inside these three spaces. I'm guessing it should be something like C(7+3-1, 2)?
    – potatoguy
    4 hours ago











  • Yes but then be careful, as you also need to permutate them to fix their order! Using the combinations with C you only decide which letters go to which box and you don't get to count their orderings. Give it a thought. If you can't make it I will further develop my answer.
    – RGS
    4 hours ago
















  • I see! So given n amount of letters besides A and E, I need to find the number of permutations of how they can be placed inside these three spaces. I'm guessing it should be something like C(7+3-1, 2)?
    – potatoguy
    4 hours ago











  • Yes but then be careful, as you also need to permutate them to fix their order! Using the combinations with C you only decide which letters go to which box and you don't get to count their orderings. Give it a thought. If you can't make it I will further develop my answer.
    – RGS
    4 hours ago















I see! So given n amount of letters besides A and E, I need to find the number of permutations of how they can be placed inside these three spaces. I'm guessing it should be something like C(7+3-1, 2)?
– potatoguy
4 hours ago





I see! So given n amount of letters besides A and E, I need to find the number of permutations of how they can be placed inside these three spaces. I'm guessing it should be something like C(7+3-1, 2)?
– potatoguy
4 hours ago













Yes but then be careful, as you also need to permutate them to fix their order! Using the combinations with C you only decide which letters go to which box and you don't get to count their orderings. Give it a thought. If you can't make it I will further develop my answer.
– RGS
4 hours ago




Yes but then be careful, as you also need to permutate them to fix their order! Using the combinations with C you only decide which letters go to which box and you don't get to count their orderings. Give it a thought. If you can't make it I will further develop my answer.
– RGS
4 hours ago










up vote
2
down vote













1-4 look fine.



5)



In exactly $frac 12$ of all permuations will A be to the left of E, and the other half it will be to the right.



$frac 9!2$



6) We have a sequence $AxE$ there are $7$ values that $x$ can be. And then think of $AxE$ as a single letter.



$7cdot 7!$






share|cite|improve this answer




















  • Your answer is quite clever. I think it would've been better, however, if you made the OP get there instead of just showing the solution :'(
    – RGS
    4 hours ago










  • Thanks for writing the explanation as well. It's important for me to know why the solution is correct. In hindsight, I really overcomplicated it in my head.
    – potatoguy
    4 hours ago






  • 1




    @RGS Thanks for having so much faith in me. It still took me some time to understand Doug's answer.
    – potatoguy
    4 hours ago














up vote
2
down vote













1-4 look fine.



5)



In exactly $frac 12$ of all permuations will A be to the left of E, and the other half it will be to the right.



$frac 9!2$



6) We have a sequence $AxE$ there are $7$ values that $x$ can be. And then think of $AxE$ as a single letter.



$7cdot 7!$






share|cite|improve this answer




















  • Your answer is quite clever. I think it would've been better, however, if you made the OP get there instead of just showing the solution :'(
    – RGS
    4 hours ago










  • Thanks for writing the explanation as well. It's important for me to know why the solution is correct. In hindsight, I really overcomplicated it in my head.
    – potatoguy
    4 hours ago






  • 1




    @RGS Thanks for having so much faith in me. It still took me some time to understand Doug's answer.
    – potatoguy
    4 hours ago












up vote
2
down vote










up vote
2
down vote









1-4 look fine.



5)



In exactly $frac 12$ of all permuations will A be to the left of E, and the other half it will be to the right.



$frac 9!2$



6) We have a sequence $AxE$ there are $7$ values that $x$ can be. And then think of $AxE$ as a single letter.



$7cdot 7!$






share|cite|improve this answer












1-4 look fine.



5)



In exactly $frac 12$ of all permuations will A be to the left of E, and the other half it will be to the right.



$frac 9!2$



6) We have a sequence $AxE$ there are $7$ values that $x$ can be. And then think of $AxE$ as a single letter.



$7cdot 7!$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Doug M

41.7k31751




41.7k31751











  • Your answer is quite clever. I think it would've been better, however, if you made the OP get there instead of just showing the solution :'(
    – RGS
    4 hours ago










  • Thanks for writing the explanation as well. It's important for me to know why the solution is correct. In hindsight, I really overcomplicated it in my head.
    – potatoguy
    4 hours ago






  • 1




    @RGS Thanks for having so much faith in me. It still took me some time to understand Doug's answer.
    – potatoguy
    4 hours ago
















  • Your answer is quite clever. I think it would've been better, however, if you made the OP get there instead of just showing the solution :'(
    – RGS
    4 hours ago










  • Thanks for writing the explanation as well. It's important for me to know why the solution is correct. In hindsight, I really overcomplicated it in my head.
    – potatoguy
    4 hours ago






  • 1




    @RGS Thanks for having so much faith in me. It still took me some time to understand Doug's answer.
    – potatoguy
    4 hours ago















Your answer is quite clever. I think it would've been better, however, if you made the OP get there instead of just showing the solution :'(
– RGS
4 hours ago




Your answer is quite clever. I think it would've been better, however, if you made the OP get there instead of just showing the solution :'(
– RGS
4 hours ago












Thanks for writing the explanation as well. It's important for me to know why the solution is correct. In hindsight, I really overcomplicated it in my head.
– potatoguy
4 hours ago




Thanks for writing the explanation as well. It's important for me to know why the solution is correct. In hindsight, I really overcomplicated it in my head.
– potatoguy
4 hours ago




1




1




@RGS Thanks for having so much faith in me. It still took me some time to understand Doug's answer.
– potatoguy
4 hours ago




@RGS Thanks for having so much faith in me. It still took me some time to understand Doug's answer.
– potatoguy
4 hours ago

















 

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