How can i find how many times a letter appears if there are only 8 positions?
$begingroup$
I'm having trouble with this question.
a appears in the first position 50388 times
b appears in the first position 31824 times and in the second 18564 times etc,
so in order to calculate a
i did $19C7$
for b: it was $18C7$ for the first position and $18C6$ for the second
for c I did $17C7$ and got 19448 but $17C6$ or $17C5$ didn't work
What am I doing wrong?
I know all the letters appear 50388 times but how can i explain this without adding each individual total
i think if i understand part a then i can do part b
combinatorics combinations
asked Dec 20 '18 at 3:54
user477465user477465
105113
$endgroup$
add a comment |
$begingroup$
I'm having trouble with this question.
a appears in the first position 50388 times
b appears in the first position 31824 times and in the second 18564 times etc,
so in order to calculate a
i did $19C7$
for b: it was $18C7$ for the first position and $18C6$ for the second
for c I did $17C7$ and got 19448 but $17C6$ or $17C5$ didn't work
What am I doing wrong?
I know all the letters appear 50388 times but how can i explain this without adding each individual total
i think if i understand part a then i can do part b
combinatorics combinations
asked Dec 20 '18 at 3:54
user477465user477465
105113
$endgroup$
add a comment |
$begingroup$
I'm having trouble with this question.
a appears in the first position 50388 times
b appears in the first position 31824 times and in the second 18564 times etc,
so in order to calculate a
i did $19C7$
for b: it was $18C7$ for the first position and $18C6$ for the second
for c I did $17C7$ and got 19448 but $17C6$ or $17C5$ didn't work
What am I doing wrong?
I know all the letters appear 50388 times but how can i explain this without adding each individual total
i think if i understand part a then i can do part b
combinatorics combinations
asked Dec 20 '18 at 3:54
user477465user477465
105113
$endgroup$
I'm having trouble with this question.
a appears in the first position 50388 times
b appears in the first position 31824 times and in the second 18564 times etc,
so in order to calculate a
i did $19C7$
for b: it was $18C7$ for the first position and $18C6$ for the second
for c I did $17C7$ and got 19448 but $17C6$ or $17C5$ didn't work
What am I doing wrong?
I know all the letters appear 50388 times but how can i explain this without adding each individual total
i think if i understand part a then i can do part b
combinatorics combinations
combinatorics combinations
asked Dec 20 '18 at 3:54
user477465user477465
105113
asked Dec 20 '18 at 3:54
user477465user477465
105113
edited Dec 20 '18 at 4:15
user477465
asked Dec 20 '18 at 3:54
user477465user477465
105113
asked Dec 20 '18 at 3:54
user477465user477465
105113
asked Dec 20 '18 at 3:54
user477465user477465
105113
105113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your set of words is all the ${20 choose 8}=125 970$ words by taking $8$ letters out of the $20$ without replacement and keeping the letters in alphabetical order. $a$ always appears in first position if it is present because nothing else comes before. $b$ appears first if $a$ is not present, so you need to choose $7$ letters out of the remaining $18$ to complete the word. If $a$ and $b$ are both present, you have to choose $6$ out of the remaining $18$ to complete the word. For $c$, if it is in first position $a$ and $b$ are missing, so you need to choose $7$ of the remaining $17$ letters to complete the word. If it is in second position, exactly one of $a$ or $b$ is present.
answered Dec 20 '18 at 4:28
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
how can i calculate the number for if c was in the second position? it would be 17C7 + something? im still confused about what i need to use in order to solve the problem
$endgroup$
– user477465
Dec 20 '18 at 4:32
$begingroup$
how many ways to choose the letter before it? how many ways to choose the letters after?
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:33
$begingroup$
for the choosing the letter before it, there are two choices and 7 positions so 7C2?
$endgroup$
– user477465
Dec 20 '18 at 4:35
$begingroup$
No, there is only one position before it. Because the letters are kept in order in the word you never select positions, you just select letters and the positions are set.
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:36
$begingroup$
so wouldn't it be 2C1 since there are two letters that could be there but only one letter that fits?
$endgroup$
– user477465
Dec 20 '18 at 4:37
|
show 10 more comments
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1 Answer
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1 Answer
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$begingroup$
Your set of words is all the ${20 choose 8}=125 970$ words by taking $8$ letters out of the $20$ without replacement and keeping the letters in alphabetical order. $a$ always appears in first position if it is present because nothing else comes before. $b$ appears first if $a$ is not present, so you need to choose $7$ letters out of the remaining $18$ to complete the word. If $a$ and $b$ are both present, you have to choose $6$ out of the remaining $18$ to complete the word. For $c$, if it is in first position $a$ and $b$ are missing, so you need to choose $7$ of the remaining $17$ letters to complete the word. If it is in second position, exactly one of $a$ or $b$ is present.
answered Dec 20 '18 at 4:28
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
how can i calculate the number for if c was in the second position? it would be 17C7 + something? im still confused about what i need to use in order to solve the problem
$endgroup$
– user477465
Dec 20 '18 at 4:32
$begingroup$
how many ways to choose the letter before it? how many ways to choose the letters after?
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:33
$begingroup$
for the choosing the letter before it, there are two choices and 7 positions so 7C2?
$endgroup$
– user477465
Dec 20 '18 at 4:35
$begingroup$
No, there is only one position before it. Because the letters are kept in order in the word you never select positions, you just select letters and the positions are set.
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:36
$begingroup$
so wouldn't it be 2C1 since there are two letters that could be there but only one letter that fits?
$endgroup$
– user477465
Dec 20 '18 at 4:37
|
show 10 more comments
$begingroup$
Your set of words is all the ${20 choose 8}=125 970$ words by taking $8$ letters out of the $20$ without replacement and keeping the letters in alphabetical order. $a$ always appears in first position if it is present because nothing else comes before. $b$ appears first if $a$ is not present, so you need to choose $7$ letters out of the remaining $18$ to complete the word. If $a$ and $b$ are both present, you have to choose $6$ out of the remaining $18$ to complete the word. For $c$, if it is in first position $a$ and $b$ are missing, so you need to choose $7$ of the remaining $17$ letters to complete the word. If it is in second position, exactly one of $a$ or $b$ is present.
answered Dec 20 '18 at 4:28
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
how can i calculate the number for if c was in the second position? it would be 17C7 + something? im still confused about what i need to use in order to solve the problem
$endgroup$
– user477465
Dec 20 '18 at 4:32
$begingroup$
how many ways to choose the letter before it? how many ways to choose the letters after?
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:33
$begingroup$
for the choosing the letter before it, there are two choices and 7 positions so 7C2?
$endgroup$
– user477465
Dec 20 '18 at 4:35
$begingroup$
No, there is only one position before it. Because the letters are kept in order in the word you never select positions, you just select letters and the positions are set.
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:36
$begingroup$
so wouldn't it be 2C1 since there are two letters that could be there but only one letter that fits?
$endgroup$
– user477465
Dec 20 '18 at 4:37
|
show 10 more comments
$begingroup$
Your set of words is all the ${20 choose 8}=125 970$ words by taking $8$ letters out of the $20$ without replacement and keeping the letters in alphabetical order. $a$ always appears in first position if it is present because nothing else comes before. $b$ appears first if $a$ is not present, so you need to choose $7$ letters out of the remaining $18$ to complete the word. If $a$ and $b$ are both present, you have to choose $6$ out of the remaining $18$ to complete the word. For $c$, if it is in first position $a$ and $b$ are missing, so you need to choose $7$ of the remaining $17$ letters to complete the word. If it is in second position, exactly one of $a$ or $b$ is present.
answered Dec 20 '18 at 4:28
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Your set of words is all the ${20 choose 8}=125 970$ words by taking $8$ letters out of the $20$ without replacement and keeping the letters in alphabetical order. $a$ always appears in first position if it is present because nothing else comes before. $b$ appears first if $a$ is not present, so you need to choose $7$ letters out of the remaining $18$ to complete the word. If $a$ and $b$ are both present, you have to choose $6$ out of the remaining $18$ to complete the word. For $c$, if it is in first position $a$ and $b$ are missing, so you need to choose $7$ of the remaining $17$ letters to complete the word. If it is in second position, exactly one of $a$ or $b$ is present.
answered Dec 20 '18 at 4:28
Ross MillikanRoss Millikan
301k24200375
answered Dec 20 '18 at 4:28
Ross MillikanRoss Millikan
301k24200375
answered Dec 20 '18 at 4:28
Ross MillikanRoss Millikan
301k24200375
answered Dec 20 '18 at 4:28
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
how can i calculate the number for if c was in the second position? it would be 17C7 + something? im still confused about what i need to use in order to solve the problem
$endgroup$
– user477465
Dec 20 '18 at 4:32
$begingroup$
how many ways to choose the letter before it? how many ways to choose the letters after?
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:33
$begingroup$
for the choosing the letter before it, there are two choices and 7 positions so 7C2?
$endgroup$
– user477465
Dec 20 '18 at 4:35
$begingroup$
No, there is only one position before it. Because the letters are kept in order in the word you never select positions, you just select letters and the positions are set.
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:36
$begingroup$
so wouldn't it be 2C1 since there are two letters that could be there but only one letter that fits?
$endgroup$
– user477465
Dec 20 '18 at 4:37
|
show 10 more comments
$begingroup$
how can i calculate the number for if c was in the second position? it would be 17C7 + something? im still confused about what i need to use in order to solve the problem
$endgroup$
– user477465
Dec 20 '18 at 4:32
$begingroup$
how many ways to choose the letter before it? how many ways to choose the letters after?
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:33
$begingroup$
for the choosing the letter before it, there are two choices and 7 positions so 7C2?
$endgroup$
– user477465
Dec 20 '18 at 4:35
$begingroup$
No, there is only one position before it. Because the letters are kept in order in the word you never select positions, you just select letters and the positions are set.
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:36
$begingroup$
so wouldn't it be 2C1 since there are two letters that could be there but only one letter that fits?
$endgroup$
– user477465
Dec 20 '18 at 4:37
$begingroup$
how can i calculate the number for if c was in the second position? it would be 17C7 + something? im still confused about what i need to use in order to solve the problem
$endgroup$
– user477465
Dec 20 '18 at 4:32
$begingroup$
how can i calculate the number for if c was in the second position? it would be 17C7 + something? im still confused about what i need to use in order to solve the problem
$endgroup$
– user477465
Dec 20 '18 at 4:32
$begingroup$
how many ways to choose the letter before it? how many ways to choose the letters after?
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:33
$begingroup$
how many ways to choose the letter before it? how many ways to choose the letters after?
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:33
$begingroup$
for the choosing the letter before it, there are two choices and 7 positions so 7C2?
$endgroup$
– user477465
Dec 20 '18 at 4:35
$begingroup$
for the choosing the letter before it, there are two choices and 7 positions so 7C2?
$endgroup$
– user477465
Dec 20 '18 at 4:35
$begingroup$
No, there is only one position before it. Because the letters are kept in order in the word you never select positions, you just select letters and the positions are set.
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:36
$begingroup$
No, there is only one position before it. Because the letters are kept in order in the word you never select positions, you just select letters and the positions are set.
$endgroup$
– Ross Millikan
Dec 20 '18 at 4:36
$begingroup$
so wouldn't it be 2C1 since there are two letters that could be there but only one letter that fits?
$endgroup$
– user477465
Dec 20 '18 at 4:37
$begingroup$
so wouldn't it be 2C1 since there are two letters that could be there but only one letter that fits?
$endgroup$
– user477465
Dec 20 '18 at 4:37
|
show 10 more comments
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