Mixing
Find the total distinct 4 letter code arrangments
$begingroup$
how many distinct 4-letter code words can be made from the letters in the word ALGEBRA?
Hello, I am currently competing in what is known as UIL Mathematics, a competitive Math event. I have tried the theory of getting the total amount of letters followed by "!" and dividing by the repeating letters followed by "!" but this didn't give the answer. I did this: 7!/2!. Please help.
combinatorics discrete-mathematics permutations
edited Jan 23 at 12:14
N. F. Taussig
44.7k103358
asked Jan 23 at 3:55
Tylor GonzalezTylor Gonzalez
12
$endgroup$
add a comment |
$begingroup$
how many distinct 4-letter code words can be made from the letters in the word ALGEBRA?
Hello, I am currently competing in what is known as UIL Mathematics, a competitive Math event. I have tried the theory of getting the total amount of letters followed by "!" and dividing by the repeating letters followed by "!" but this didn't give the answer. I did this: 7!/2!. Please help.
combinatorics discrete-mathematics permutations
edited Jan 23 at 12:14
N. F. Taussig
44.7k103358
asked Jan 23 at 3:55
Tylor GonzalezTylor Gonzalez
12
$endgroup$
1
$begingroup$
Your answer counts the number of ways to arrange ALL of the letters into a word. What we need is to find the number of ways we pick only $4$ letter words. Perhaps the hint given by JMoravitz will point you in the right direction
$endgroup$
– WaveX
Jan 23 at 4:08
add a comment |
$begingroup$
how many distinct 4-letter code words can be made from the letters in the word ALGEBRA?
Hello, I am currently competing in what is known as UIL Mathematics, a competitive Math event. I have tried the theory of getting the total amount of letters followed by "!" and dividing by the repeating letters followed by "!" but this didn't give the answer. I did this: 7!/2!. Please help.
combinatorics discrete-mathematics permutations
edited Jan 23 at 12:14
N. F. Taussig
44.7k103358
asked Jan 23 at 3:55
Tylor GonzalezTylor Gonzalez
12
$endgroup$
how many distinct 4-letter code words can be made from the letters in the word ALGEBRA?
Hello, I am currently competing in what is known as UIL Mathematics, a competitive Math event. I have tried the theory of getting the total amount of letters followed by "!" and dividing by the repeating letters followed by "!" but this didn't give the answer. I did this: 7!/2!. Please help.
combinatorics discrete-mathematics permutations
combinatorics discrete-mathematics permutations
edited Jan 23 at 12:14
N. F. Taussig
44.7k103358
asked Jan 23 at 3:55
Tylor GonzalezTylor Gonzalez
12
edited Jan 23 at 12:14
N. F. Taussig
44.7k103358
asked Jan 23 at 3:55
Tylor GonzalezTylor Gonzalez
12
edited Jan 23 at 12:14
N. F. Taussig
44.7k103358
edited Jan 23 at 12:14
N. F. Taussig
44.7k103358
edited Jan 23 at 12:14
N. F. Taussig
44.7k103358
44.7k103358
asked Jan 23 at 3:55
Tylor GonzalezTylor Gonzalez
12
asked Jan 23 at 3:55
Tylor GonzalezTylor Gonzalez
12
asked Jan 23 at 3:55
Tylor GonzalezTylor Gonzalez
12
12
1
$begingroup$
Your answer counts the number of ways to arrange ALL of the letters into a word. What we need is to find the number of ways we pick only $4$ letter words. Perhaps the hint given by JMoravitz will point you in the right direction
$endgroup$
– WaveX
Jan 23 at 4:08
add a comment |
1
$begingroup$
Your answer counts the number of ways to arrange ALL of the letters into a word. What we need is to find the number of ways we pick only $4$ letter words. Perhaps the hint given by JMoravitz will point you in the right direction
$endgroup$
– WaveX
Jan 23 at 4:08
1
1
$begingroup$
Your answer counts the number of ways to arrange ALL of the letters into a word. What we need is to find the number of ways we pick only $4$ letter words. Perhaps the hint given by JMoravitz will point you in the right direction
$endgroup$
– WaveX
Jan 23 at 4:08
$begingroup$
Your answer counts the number of ways to arrange ALL of the letters into a word. What we need is to find the number of ways we pick only $4$ letter words. Perhaps the hint given by JMoravitz will point you in the right direction
$endgroup$
– WaveX
Jan 23 at 4:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
How many ways can you choose a four letter code using letters from the word ALGEBR
The above codes will all have zero or one A.
Now, how many ways can you choose a four letter code where there are exactly two A's being used and the remaining two letters are taken from LGEBR.
answered Jan 23 at 4:07
JMoravitzJMoravitz
48.5k33987
$endgroup$
add a comment |
$begingroup$
Following @JMoravitz's hint, there should be
$P^6_4+P^5_2cdot C^4_2=360+20cdot6=480$.
answered Jan 23 at 4:45
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint:
How many ways can you choose a four letter code using letters from the word ALGEBR
The above codes will all have zero or one A.
Now, how many ways can you choose a four letter code where there are exactly two A's being used and the remaining two letters are taken from LGEBR.
answered Jan 23 at 4:07
JMoravitzJMoravitz
48.5k33987
$endgroup$
add a comment |
$begingroup$
Hint:
How many ways can you choose a four letter code using letters from the word ALGEBR
The above codes will all have zero or one A.
Now, how many ways can you choose a four letter code where there are exactly two A's being used and the remaining two letters are taken from LGEBR.
answered Jan 23 at 4:07
JMoravitzJMoravitz
48.5k33987
$endgroup$
add a comment |
$begingroup$
Hint:
How many ways can you choose a four letter code using letters from the word ALGEBR
The above codes will all have zero or one A.
Now, how many ways can you choose a four letter code where there are exactly two A's being used and the remaining two letters are taken from LGEBR.
answered Jan 23 at 4:07
JMoravitzJMoravitz
48.5k33987
$endgroup$
Hint:
How many ways can you choose a four letter code using letters from the word ALGEBR
The above codes will all have zero or one A.
Now, how many ways can you choose a four letter code where there are exactly two A's being used and the remaining two letters are taken from LGEBR.
answered Jan 23 at 4:07
JMoravitzJMoravitz
48.5k33987
answered Jan 23 at 4:07
JMoravitzJMoravitz
48.5k33987
answered Jan 23 at 4:07
JMoravitzJMoravitz
48.5k33987
answered Jan 23 at 4:07
JMoravitzJMoravitz
48.5k33987
48.5k33987
add a comment |
add a comment |
$begingroup$
Following @JMoravitz's hint, there should be
$P^6_4+P^5_2cdot C^4_2=360+20cdot6=480$.
answered Jan 23 at 4:45
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Following @JMoravitz's hint, there should be
$P^6_4+P^5_2cdot C^4_2=360+20cdot6=480$.
answered Jan 23 at 4:45
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Following @JMoravitz's hint, there should be
$P^6_4+P^5_2cdot C^4_2=360+20cdot6=480$.
answered Jan 23 at 4:45
Chris CusterChris Custer
14.2k3827
$endgroup$
Following @JMoravitz's hint, there should be
$P^6_4+P^5_2cdot C^4_2=360+20cdot6=480$.
answered Jan 23 at 4:45
Chris CusterChris Custer
14.2k3827
answered Jan 23 at 4:45
Chris CusterChris Custer
14.2k3827
answered Jan 23 at 4:45
Chris CusterChris Custer
14.2k3827
answered Jan 23 at 4:45
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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1
$begingroup$
Your answer counts the number of ways to arrange ALL of the letters into a word. What we need is to find the number of ways we pick only $4$ letter words. Perhaps the hint given by JMoravitz will point you in the right direction
$endgroup$
– WaveX
Jan 23 at 4:08