Permutations Alphabet
how many ways are there to arrange 10 letters taken from the alphabet a-z such that:
a) a is not included
b) z is included
c) both a and z are included
I believe for the first one it should be 25!/15!, but would both b and c be
26!/16!?
discrete-mathematics permutations
asked Nov 27 '18 at 15:06
Charlie
227
add a comment |
how many ways are there to arrange 10 letters taken from the alphabet a-z such that:
a) a is not included
b) z is included
c) both a and z are included
I believe for the first one it should be 25!/15!, but would both b and c be
26!/16!?
discrete-mathematics permutations
asked Nov 27 '18 at 15:06
Charlie
227
b) $frac{25!*10}{16!}$ c) $frac{24!*9*10}{17!}$ ?
– mathnoob
Nov 27 '18 at 15:11
Ok is this correct now?
– mathnoob
Nov 27 '18 at 15:16
1
c) $frac{24!*9*10}{16!}$
– mathnoob
Nov 27 '18 at 15:18
Yes @mathnoob that's correct, without any doubt
– Akash Roy
Nov 27 '18 at 15:20
For b) first I get all permutations of $9$ letters not including $z$, then I have to insert $z$ into the permutations, there are $10$ places to put $z$, hence multiply by $10$. For c) the reasoning is the same.
– mathnoob
Nov 27 '18 at 15:35
add a comment |
how many ways are there to arrange 10 letters taken from the alphabet a-z such that:
a) a is not included
b) z is included
c) both a and z are included
I believe for the first one it should be 25!/15!, but would both b and c be
26!/16!?
discrete-mathematics permutations
asked Nov 27 '18 at 15:06
Charlie
227
how many ways are there to arrange 10 letters taken from the alphabet a-z such that:
a) a is not included
b) z is included
c) both a and z are included
I believe for the first one it should be 25!/15!, but would both b and c be
26!/16!?
discrete-mathematics permutations
discrete-mathematics permutations
asked Nov 27 '18 at 15:06
Charlie
227
asked Nov 27 '18 at 15:06
Charlie
227
edited Nov 27 '18 at 15:20
asked Nov 27 '18 at 15:06
Charlie
227
asked Nov 27 '18 at 15:06
Charlie
227
asked Nov 27 '18 at 15:06
Charlie
227
227
b) $frac{25!*10}{16!}$ c) $frac{24!*9*10}{17!}$ ?
– mathnoob
Nov 27 '18 at 15:11
Ok is this correct now?
– mathnoob
Nov 27 '18 at 15:16
1
c) $frac{24!*9*10}{16!}$
– mathnoob
Nov 27 '18 at 15:18
Yes @mathnoob that's correct, without any doubt
– Akash Roy
Nov 27 '18 at 15:20
For b) first I get all permutations of $9$ letters not including $z$, then I have to insert $z$ into the permutations, there are $10$ places to put $z$, hence multiply by $10$. For c) the reasoning is the same.
– mathnoob
Nov 27 '18 at 15:35
add a comment |
b) $frac{25!*10}{16!}$ c) $frac{24!*9*10}{17!}$ ?
– mathnoob
Nov 27 '18 at 15:11
Ok is this correct now?
– mathnoob
Nov 27 '18 at 15:16
1
c) $frac{24!*9*10}{16!}$
– mathnoob
Nov 27 '18 at 15:18
Yes @mathnoob that's correct, without any doubt
– Akash Roy
Nov 27 '18 at 15:20
For b) first I get all permutations of $9$ letters not including $z$, then I have to insert $z$ into the permutations, there are $10$ places to put $z$, hence multiply by $10$. For c) the reasoning is the same.
– mathnoob
Nov 27 '18 at 15:35
b) $frac{25!*10}{16!}$ c) $frac{24!*9*10}{17!}$ ?
– mathnoob
Nov 27 '18 at 15:11
b) $frac{25!*10}{16!}$ c) $frac{24!*9*10}{17!}$ ?
– mathnoob
Nov 27 '18 at 15:11
Ok is this correct now?
– mathnoob
Nov 27 '18 at 15:16
Ok is this correct now?
– mathnoob
Nov 27 '18 at 15:16
1
1
c) $frac{24!*9*10}{16!}$
– mathnoob
Nov 27 '18 at 15:18
c) $frac{24!*9*10}{16!}$
– mathnoob
Nov 27 '18 at 15:18
Yes @mathnoob that's correct, without any doubt
– Akash Roy
Nov 27 '18 at 15:20
Yes @mathnoob that's correct, without any doubt
– Akash Roy
Nov 27 '18 at 15:20
For b) first I get all permutations of $9$ letters not including $z$, then I have to insert $z$ into the permutations, there are $10$ places to put $z$, hence multiply by $10$. For c) the reasoning is the same.
– mathnoob
Nov 27 '18 at 15:35
For b) first I get all permutations of $9$ letters not including $z$, then I have to insert $z$ into the permutations, there are $10$ places to put $z$, hence multiply by $10$. For c) the reasoning is the same.
– mathnoob
Nov 27 '18 at 15:35
add a comment |
1 Answer
1
active
oldest
votes
For second case it should be $25choose 9$$×10!$
In this case , I have chosen $9$ letters out of $25$ since $Z$ is already included. After that I arranged the letters
For third case it should be $24choose 8$$×10!$.
In this case , I have chosen $8$ letters out of $24$ since $Z$ and $A$ are already included. After that I arranged the letters
Note: $nchoose r$ stands for combination.
answered Nov 27 '18 at 15:10
Akash Roy
1
I have been using the permutations formula p(n,r) = n!/(n-r)! So I am not quite sure why one would use combinations? I thought this was a permutation question because the order matters.
– Charlie
Nov 27 '18 at 15:17
2
For b) first you choose $9$ letters out of $25$ because you already choose $z$. Then you permute $10$ letters so you multiply by $10!$. For c) first you choose $8$ letters out of $24$ because you already choose $a$ and $z$, then you permute $10$ letters.
– mathnoob
Nov 27 '18 at 15:21
2
Yes, elaborate it a little , it will help other to understand
– Cloud JR
Nov 27 '18 at 15:30
1
Please check the elaborated answer and let me know if more is to be added.
– Akash Roy
Nov 27 '18 at 15:31
1
Yes man of course.
– Akash Roy
Nov 27 '18 at 15:33
|
show 16 more comments
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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oldest
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For second case it should be $25choose 9$$×10!$
In this case , I have chosen $9$ letters out of $25$ since $Z$ is already included. After that I arranged the letters
For third case it should be $24choose 8$$×10!$.
In this case , I have chosen $8$ letters out of $24$ since $Z$ and $A$ are already included. After that I arranged the letters
Note: $nchoose r$ stands for combination.
answered Nov 27 '18 at 15:10
Akash Roy
1
I have been using the permutations formula p(n,r) = n!/(n-r)! So I am not quite sure why one would use combinations? I thought this was a permutation question because the order matters.
– Charlie
Nov 27 '18 at 15:17
2
For b) first you choose $9$ letters out of $25$ because you already choose $z$. Then you permute $10$ letters so you multiply by $10!$. For c) first you choose $8$ letters out of $24$ because you already choose $a$ and $z$, then you permute $10$ letters.
– mathnoob
Nov 27 '18 at 15:21
2
Yes, elaborate it a little , it will help other to understand
– Cloud JR
Nov 27 '18 at 15:30
1
Please check the elaborated answer and let me know if more is to be added.
– Akash Roy
Nov 27 '18 at 15:31
1
Yes man of course.
– Akash Roy
Nov 27 '18 at 15:33
|
show 16 more comments
For second case it should be $25choose 9$$×10!$
In this case , I have chosen $9$ letters out of $25$ since $Z$ is already included. After that I arranged the letters
For third case it should be $24choose 8$$×10!$.
In this case , I have chosen $8$ letters out of $24$ since $Z$ and $A$ are already included. After that I arranged the letters
Note: $nchoose r$ stands for combination.
answered Nov 27 '18 at 15:10
Akash Roy
1
I have been using the permutations formula p(n,r) = n!/(n-r)! So I am not quite sure why one would use combinations? I thought this was a permutation question because the order matters.
– Charlie
Nov 27 '18 at 15:17
2
For b) first you choose $9$ letters out of $25$ because you already choose $z$. Then you permute $10$ letters so you multiply by $10!$. For c) first you choose $8$ letters out of $24$ because you already choose $a$ and $z$, then you permute $10$ letters.
– mathnoob
Nov 27 '18 at 15:21
2
Yes, elaborate it a little , it will help other to understand
– Cloud JR
Nov 27 '18 at 15:30
1
Please check the elaborated answer and let me know if more is to be added.
– Akash Roy
Nov 27 '18 at 15:31
1
Yes man of course.
– Akash Roy
Nov 27 '18 at 15:33
|
show 16 more comments
For second case it should be $25choose 9$$×10!$
In this case , I have chosen $9$ letters out of $25$ since $Z$ is already included. After that I arranged the letters
For third case it should be $24choose 8$$×10!$.
In this case , I have chosen $8$ letters out of $24$ since $Z$ and $A$ are already included. After that I arranged the letters
Note: $nchoose r$ stands for combination.
answered Nov 27 '18 at 15:10
Akash Roy
1
For second case it should be $25choose 9$$×10!$
In this case , I have chosen $9$ letters out of $25$ since $Z$ is already included. After that I arranged the letters
For third case it should be $24choose 8$$×10!$.
In this case , I have chosen $8$ letters out of $24$ since $Z$ and $A$ are already included. After that I arranged the letters
Note: $nchoose r$ stands for combination.
answered Nov 27 '18 at 15:10
Akash Roy
1
edited Nov 27 '18 at 15:30
answered Nov 27 '18 at 15:10
Akash Roy
1
answered Nov 27 '18 at 15:10
Akash Roy
1
answered Nov 27 '18 at 15:10
Akash Roy
1
1
I have been using the permutations formula p(n,r) = n!/(n-r)! So I am not quite sure why one would use combinations? I thought this was a permutation question because the order matters.
– Charlie
Nov 27 '18 at 15:17
2
For b) first you choose $9$ letters out of $25$ because you already choose $z$. Then you permute $10$ letters so you multiply by $10!$. For c) first you choose $8$ letters out of $24$ because you already choose $a$ and $z$, then you permute $10$ letters.
– mathnoob
Nov 27 '18 at 15:21
2
Yes, elaborate it a little , it will help other to understand
– Cloud JR
Nov 27 '18 at 15:30
1
Please check the elaborated answer and let me know if more is to be added.
– Akash Roy
Nov 27 '18 at 15:31
1
Yes man of course.
– Akash Roy
Nov 27 '18 at 15:33
|
show 16 more comments
I have been using the permutations formula p(n,r) = n!/(n-r)! So I am not quite sure why one would use combinations? I thought this was a permutation question because the order matters.
– Charlie
Nov 27 '18 at 15:17
2
For b) first you choose $9$ letters out of $25$ because you already choose $z$. Then you permute $10$ letters so you multiply by $10!$. For c) first you choose $8$ letters out of $24$ because you already choose $a$ and $z$, then you permute $10$ letters.
– mathnoob
Nov 27 '18 at 15:21
2
Yes, elaborate it a little , it will help other to understand
– Cloud JR
Nov 27 '18 at 15:30
1
Please check the elaborated answer and let me know if more is to be added.
– Akash Roy
Nov 27 '18 at 15:31
1
Yes man of course.
– Akash Roy
Nov 27 '18 at 15:33
I have been using the permutations formula p(n,r) = n!/(n-r)! So I am not quite sure why one would use combinations? I thought this was a permutation question because the order matters.
– Charlie
Nov 27 '18 at 15:17
I have been using the permutations formula p(n,r) = n!/(n-r)! So I am not quite sure why one would use combinations? I thought this was a permutation question because the order matters.
– Charlie
Nov 27 '18 at 15:17
2
2
For b) first you choose $9$ letters out of $25$ because you already choose $z$. Then you permute $10$ letters so you multiply by $10!$. For c) first you choose $8$ letters out of $24$ because you already choose $a$ and $z$, then you permute $10$ letters.
– mathnoob
Nov 27 '18 at 15:21
For b) first you choose $9$ letters out of $25$ because you already choose $z$. Then you permute $10$ letters so you multiply by $10!$. For c) first you choose $8$ letters out of $24$ because you already choose $a$ and $z$, then you permute $10$ letters.
– mathnoob
Nov 27 '18 at 15:21
2
2
Yes, elaborate it a little , it will help other to understand
– Cloud JR
Nov 27 '18 at 15:30
Yes, elaborate it a little , it will help other to understand
– Cloud JR
Nov 27 '18 at 15:30
1
1
Please check the elaborated answer and let me know if more is to be added.
– Akash Roy
Nov 27 '18 at 15:31
Please check the elaborated answer and let me know if more is to be added.
– Akash Roy
Nov 27 '18 at 15:31
1
1
Yes man of course.
– Akash Roy
Nov 27 '18 at 15:33
Yes man of course.
– Akash Roy
Nov 27 '18 at 15:33
|
show 16 more comments
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b) $frac{25!*10}{16!}$ c) $frac{24!*9*10}{17!}$ ?
– mathnoob
Nov 27 '18 at 15:11
Ok is this correct now?
– mathnoob
Nov 27 '18 at 15:16
1
c) $frac{24!*9*10}{16!}$
– mathnoob
Nov 27 '18 at 15:18
Yes @mathnoob that's correct, without any doubt
– Akash Roy
Nov 27 '18 at 15:20
For b) first I get all permutations of $9$ letters not including $z$, then I have to insert $z$ into the permutations, there are $10$ places to put $z$, hence multiply by $10$. For c) the reasoning is the same.
– mathnoob
Nov 27 '18 at 15:35