Mixing
Distinct sequences of characters
$begingroup$
I'm new to combinatorial calculus. I have a question for you that I am trying to answer.
How many distinct sequences can be formed using exactly 5 times the letter A, 6 times the letter B and 7 times the letter C?
How many can be formed by choosing an additional letter of the alphabet from the remaining 23?
I asked the first question with the following calculation: $(18!)/(5!6!7!) $ (sorry for the fraction but I have not found a way to write it correctly).
For the second question, I have my doubts as to how to resolve it. Do I have to approach this question thinking that I only have one more place in the sequence, or do I have to understand how many ways I can occupy one of the 19 places by selecting a letter from 23?
combinatorics
asked Jan 15 at 19:20
PCNFPCNF
1338
$endgroup$
add a comment |
$begingroup$
I'm new to combinatorial calculus. I have a question for you that I am trying to answer.
How many distinct sequences can be formed using exactly 5 times the letter A, 6 times the letter B and 7 times the letter C?
How many can be formed by choosing an additional letter of the alphabet from the remaining 23?
I asked the first question with the following calculation: $(18!)/(5!6!7!) $ (sorry for the fraction but I have not found a way to write it correctly).
For the second question, I have my doubts as to how to resolve it. Do I have to approach this question thinking that I only have one more place in the sequence, or do I have to understand how many ways I can occupy one of the 19 places by selecting a letter from 23?
combinatorics
asked Jan 15 at 19:20
PCNFPCNF
1338
$endgroup$
$begingroup$
Visit this page for information on how to properly typeset mathematics here using MathJax and $LaTeX$ commands. Your answer for the first part of the question is correct. Now... for the second part of the question, approach it in two steps. First, count how many sequences there are using five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$. Next, find how many ways there are to replace the $heartsuit$ with a letter and then apply multiplication principle.
$endgroup$
– JMoravitz
Jan 15 at 19:30
$begingroup$
If your question is just how to interpret the question, be it we are only wanting to count sequences of the form (AAA..BB..CC..)(D), (AAA..BB..CC..)(E), ... where the new letter is only appended at the end, that is not how I interpret it. I interpret it as the new letter can be anywhere, not just at the end, soAAXABBBCCCBBBCCCCAAwould be a valid sequence which we wish to count for example.
$endgroup$
– JMoravitz
Jan 15 at 19:33
$begingroup$
@JMoravitz I am also of the same opinion as you and I think that the additional letter can be placed anywhere in the sequence. My question is aimed at trying to understand if I answered the first question well and understand how to answer the second. EDIT: because of the slow network I have not seen your answer divided into two parts. Thank you for the suggestion, I will try to proceed as you suggested.
$endgroup$
– PCNF
Jan 15 at 19:36
add a comment |
$begingroup$
I'm new to combinatorial calculus. I have a question for you that I am trying to answer.
How many distinct sequences can be formed using exactly 5 times the letter A, 6 times the letter B and 7 times the letter C?
How many can be formed by choosing an additional letter of the alphabet from the remaining 23?
I asked the first question with the following calculation: $(18!)/(5!6!7!) $ (sorry for the fraction but I have not found a way to write it correctly).
For the second question, I have my doubts as to how to resolve it. Do I have to approach this question thinking that I only have one more place in the sequence, or do I have to understand how many ways I can occupy one of the 19 places by selecting a letter from 23?
combinatorics
asked Jan 15 at 19:20
PCNFPCNF
1338
$endgroup$
I'm new to combinatorial calculus. I have a question for you that I am trying to answer.
How many distinct sequences can be formed using exactly 5 times the letter A, 6 times the letter B and 7 times the letter C?
How many can be formed by choosing an additional letter of the alphabet from the remaining 23?
I asked the first question with the following calculation: $(18!)/(5!6!7!) $ (sorry for the fraction but I have not found a way to write it correctly).
For the second question, I have my doubts as to how to resolve it. Do I have to approach this question thinking that I only have one more place in the sequence, or do I have to understand how many ways I can occupy one of the 19 places by selecting a letter from 23?
combinatorics
combinatorics
asked Jan 15 at 19:20
PCNFPCNF
1338
asked Jan 15 at 19:20
PCNFPCNF
1338
asked Jan 15 at 19:20
PCNFPCNF
1338
asked Jan 15 at 19:20
PCNFPCNF
1338
asked Jan 15 at 19:20
PCNFPCNF
1338
1338
$begingroup$
Visit this page for information on how to properly typeset mathematics here using MathJax and $LaTeX$ commands. Your answer for the first part of the question is correct. Now... for the second part of the question, approach it in two steps. First, count how many sequences there are using five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$. Next, find how many ways there are to replace the $heartsuit$ with a letter and then apply multiplication principle.
$endgroup$
– JMoravitz
Jan 15 at 19:30
$begingroup$
If your question is just how to interpret the question, be it we are only wanting to count sequences of the form (AAA..BB..CC..)(D), (AAA..BB..CC..)(E), ... where the new letter is only appended at the end, that is not how I interpret it. I interpret it as the new letter can be anywhere, not just at the end, soAAXABBBCCCBBBCCCCAAwould be a valid sequence which we wish to count for example.
$endgroup$
– JMoravitz
Jan 15 at 19:33
$begingroup$
@JMoravitz I am also of the same opinion as you and I think that the additional letter can be placed anywhere in the sequence. My question is aimed at trying to understand if I answered the first question well and understand how to answer the second. EDIT: because of the slow network I have not seen your answer divided into two parts. Thank you for the suggestion, I will try to proceed as you suggested.
$endgroup$
– PCNF
Jan 15 at 19:36
add a comment |
$begingroup$
Visit this page for information on how to properly typeset mathematics here using MathJax and $LaTeX$ commands. Your answer for the first part of the question is correct. Now... for the second part of the question, approach it in two steps. First, count how many sequences there are using five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$. Next, find how many ways there are to replace the $heartsuit$ with a letter and then apply multiplication principle.
$endgroup$
– JMoravitz
Jan 15 at 19:30
$begingroup$
If your question is just how to interpret the question, be it we are only wanting to count sequences of the form (AAA..BB..CC..)(D), (AAA..BB..CC..)(E), ... where the new letter is only appended at the end, that is not how I interpret it. I interpret it as the new letter can be anywhere, not just at the end, soAAXABBBCCCBBBCCCCAAwould be a valid sequence which we wish to count for example.
$endgroup$
– JMoravitz
Jan 15 at 19:33
$begingroup$
@JMoravitz I am also of the same opinion as you and I think that the additional letter can be placed anywhere in the sequence. My question is aimed at trying to understand if I answered the first question well and understand how to answer the second. EDIT: because of the slow network I have not seen your answer divided into two parts. Thank you for the suggestion, I will try to proceed as you suggested.
$endgroup$
– PCNF
Jan 15 at 19:36
$begingroup$
Visit this page for information on how to properly typeset mathematics here using MathJax and $LaTeX$ commands. Your answer for the first part of the question is correct. Now... for the second part of the question, approach it in two steps. First, count how many sequences there are using five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$. Next, find how many ways there are to replace the $heartsuit$ with a letter and then apply multiplication principle.
$endgroup$
– JMoravitz
Jan 15 at 19:30
$begingroup$
Visit this page for information on how to properly typeset mathematics here using MathJax and $LaTeX$ commands. Your answer for the first part of the question is correct. Now... for the second part of the question, approach it in two steps. First, count how many sequences there are using five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$. Next, find how many ways there are to replace the $heartsuit$ with a letter and then apply multiplication principle.
$endgroup$
– JMoravitz
Jan 15 at 19:30
$begingroup$
If your question is just how to interpret the question, be it we are only wanting to count sequences of the form (AAA..BB..CC..)(D), (AAA..BB..CC..)(E), ... where the new letter is only appended at the end, that is not how I interpret it. I interpret it as the new letter can be anywhere, not just at the end, so
AAXABBBCCCBBBCCCCAA would be a valid sequence which we wish to count for example.$endgroup$
– JMoravitz
Jan 15 at 19:33
$begingroup$
If your question is just how to interpret the question, be it we are only wanting to count sequences of the form (AAA..BB..CC..)(D), (AAA..BB..CC..)(E), ... where the new letter is only appended at the end, that is not how I interpret it. I interpret it as the new letter can be anywhere, not just at the end, so
AAXABBBCCCBBBCCCCAA would be a valid sequence which we wish to count for example.$endgroup$
– JMoravitz
Jan 15 at 19:33
$begingroup$
@JMoravitz I am also of the same opinion as you and I think that the additional letter can be placed anywhere in the sequence. My question is aimed at trying to understand if I answered the first question well and understand how to answer the second. EDIT: because of the slow network I have not seen your answer divided into two parts. Thank you for the suggestion, I will try to proceed as you suggested.
$endgroup$
– PCNF
Jan 15 at 19:36
$begingroup$
@JMoravitz I am also of the same opinion as you and I think that the additional letter can be placed anywhere in the sequence. My question is aimed at trying to understand if I answered the first question well and understand how to answer the second. EDIT: because of the slow network I have not seen your answer divided into two parts. Thank you for the suggestion, I will try to proceed as you suggested.
$endgroup$
– PCNF
Jan 15 at 19:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First answer is correct and is the multinomial coefficient $binom{18}{5,6,7}=dfrac{18!}{5!6!7!}$.
The second answer can be found a number of ways, most introductory ways will involve multiplication principle and binomial or multinomial coefficients in some way.
As alluded to in comments above, one way would be to arrange five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$ into a line and then pick which letter the $heartsuit$ is replaced with.
$binom{19}{5,6,7,1}cdot 23=frac{19!}{5!6!7!1!}cdot 23$
This is equivalent to first arranging five $A$'s, six $B$'s, seven $C$'s in a line and then picking a spot to insert a $heartsuit$ inbetween two characters or the front or back and then choosing which letter $heartsuit$ represents.
$binom{18}{5,6,7}cdot 19cdot 23$
You could completely reverse the order of steps as well in this case by first picking what the new letter is, then picking where it goes, then picking where the $A$'s go, then where the $B$'s, finally the $C$'s
$23cdot 19cdot binom{18}{5}cdot binom{13}{6}cdot 1$
In the end, just check that each outcome you wish to count is counted, and further each outcome is counted only once and not multiple times each.
answered Jan 15 at 19:44
JMoravitzJMoravitz
47.9k33886
$endgroup$
add a comment |
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1 Answer
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$begingroup$
First answer is correct and is the multinomial coefficient $binom{18}{5,6,7}=dfrac{18!}{5!6!7!}$.
The second answer can be found a number of ways, most introductory ways will involve multiplication principle and binomial or multinomial coefficients in some way.
As alluded to in comments above, one way would be to arrange five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$ into a line and then pick which letter the $heartsuit$ is replaced with.
$binom{19}{5,6,7,1}cdot 23=frac{19!}{5!6!7!1!}cdot 23$
This is equivalent to first arranging five $A$'s, six $B$'s, seven $C$'s in a line and then picking a spot to insert a $heartsuit$ inbetween two characters or the front or back and then choosing which letter $heartsuit$ represents.
$binom{18}{5,6,7}cdot 19cdot 23$
You could completely reverse the order of steps as well in this case by first picking what the new letter is, then picking where it goes, then picking where the $A$'s go, then where the $B$'s, finally the $C$'s
$23cdot 19cdot binom{18}{5}cdot binom{13}{6}cdot 1$
In the end, just check that each outcome you wish to count is counted, and further each outcome is counted only once and not multiple times each.
answered Jan 15 at 19:44
JMoravitzJMoravitz
47.9k33886
$endgroup$
add a comment |
$begingroup$
First answer is correct and is the multinomial coefficient $binom{18}{5,6,7}=dfrac{18!}{5!6!7!}$.
The second answer can be found a number of ways, most introductory ways will involve multiplication principle and binomial or multinomial coefficients in some way.
As alluded to in comments above, one way would be to arrange five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$ into a line and then pick which letter the $heartsuit$ is replaced with.
$binom{19}{5,6,7,1}cdot 23=frac{19!}{5!6!7!1!}cdot 23$
This is equivalent to first arranging five $A$'s, six $B$'s, seven $C$'s in a line and then picking a spot to insert a $heartsuit$ inbetween two characters or the front or back and then choosing which letter $heartsuit$ represents.
$binom{18}{5,6,7}cdot 19cdot 23$
You could completely reverse the order of steps as well in this case by first picking what the new letter is, then picking where it goes, then picking where the $A$'s go, then where the $B$'s, finally the $C$'s
$23cdot 19cdot binom{18}{5}cdot binom{13}{6}cdot 1$
In the end, just check that each outcome you wish to count is counted, and further each outcome is counted only once and not multiple times each.
answered Jan 15 at 19:44
JMoravitzJMoravitz
47.9k33886
$endgroup$
add a comment |
$begingroup$
First answer is correct and is the multinomial coefficient $binom{18}{5,6,7}=dfrac{18!}{5!6!7!}$.
The second answer can be found a number of ways, most introductory ways will involve multiplication principle and binomial or multinomial coefficients in some way.
As alluded to in comments above, one way would be to arrange five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$ into a line and then pick which letter the $heartsuit$ is replaced with.
$binom{19}{5,6,7,1}cdot 23=frac{19!}{5!6!7!1!}cdot 23$
This is equivalent to first arranging five $A$'s, six $B$'s, seven $C$'s in a line and then picking a spot to insert a $heartsuit$ inbetween two characters or the front or back and then choosing which letter $heartsuit$ represents.
$binom{18}{5,6,7}cdot 19cdot 23$
You could completely reverse the order of steps as well in this case by first picking what the new letter is, then picking where it goes, then picking where the $A$'s go, then where the $B$'s, finally the $C$'s
$23cdot 19cdot binom{18}{5}cdot binom{13}{6}cdot 1$
In the end, just check that each outcome you wish to count is counted, and further each outcome is counted only once and not multiple times each.
answered Jan 15 at 19:44
JMoravitzJMoravitz
47.9k33886
$endgroup$
First answer is correct and is the multinomial coefficient $binom{18}{5,6,7}=dfrac{18!}{5!6!7!}$.
The second answer can be found a number of ways, most introductory ways will involve multiplication principle and binomial or multinomial coefficients in some way.
As alluded to in comments above, one way would be to arrange five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$ into a line and then pick which letter the $heartsuit$ is replaced with.
$binom{19}{5,6,7,1}cdot 23=frac{19!}{5!6!7!1!}cdot 23$
This is equivalent to first arranging five $A$'s, six $B$'s, seven $C$'s in a line and then picking a spot to insert a $heartsuit$ inbetween two characters or the front or back and then choosing which letter $heartsuit$ represents.
$binom{18}{5,6,7}cdot 19cdot 23$
You could completely reverse the order of steps as well in this case by first picking what the new letter is, then picking where it goes, then picking where the $A$'s go, then where the $B$'s, finally the $C$'s
$23cdot 19cdot binom{18}{5}cdot binom{13}{6}cdot 1$
In the end, just check that each outcome you wish to count is counted, and further each outcome is counted only once and not multiple times each.
answered Jan 15 at 19:44
JMoravitzJMoravitz
47.9k33886
answered Jan 15 at 19:44
JMoravitzJMoravitz
47.9k33886
answered Jan 15 at 19:44
JMoravitzJMoravitz
47.9k33886
answered Jan 15 at 19:44
JMoravitzJMoravitz
47.9k33886
47.9k33886
add a comment |
add a comment |
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$begingroup$
Visit this page for information on how to properly typeset mathematics here using MathJax and $LaTeX$ commands. Your answer for the first part of the question is correct. Now... for the second part of the question, approach it in two steps. First, count how many sequences there are using five $A$'s, six $B$'s, seven $C$'s, and one $heartsuit$. Next, find how many ways there are to replace the $heartsuit$ with a letter and then apply multiplication principle.
$endgroup$
– JMoravitz
Jan 15 at 19:30
$begingroup$
If your question is just how to interpret the question, be it we are only wanting to count sequences of the form (AAA..BB..CC..)(D), (AAA..BB..CC..)(E), ... where the new letter is only appended at the end, that is not how I interpret it. I interpret it as the new letter can be anywhere, not just at the end, so
AAXABBBCCCBBBCCCCAAwould be a valid sequence which we wish to count for example.$endgroup$
– JMoravitz
Jan 15 at 19:33
$begingroup$
@JMoravitz I am also of the same opinion as you and I think that the additional letter can be placed anywhere in the sequence. My question is aimed at trying to understand if I answered the first question well and understand how to answer the second. EDIT: because of the slow network I have not seen your answer divided into two parts. Thank you for the suggestion, I will try to proceed as you suggested.
$endgroup$
– PCNF
Jan 15 at 19:36