how many permutation of the letters PQRSTUVWX contain the string VWX or WX
up vote
-1
down vote
favorite
I know how to calculate each task like the following
7! + 8!
but I am not sure how to subtract the overlapping part
usually, it will the (or) parts is between distinct string, but in this example, they share 2 letters
how do I solve this?
discrete-mathematics permutations
asked Nov 16 at 15:36
Nour M Al-S
1
|
show 1 more comment
up vote
-1
down vote
favorite
I know how to calculate each task like the following
7! + 8!
but I am not sure how to subtract the overlapping part
usually, it will the (or) parts is between distinct string, but in this example, they share 2 letters
how do I solve this?
discrete-mathematics permutations
asked Nov 16 at 15:36
Nour M Al-S
1
3
If the permutation contains $VWX$ then it automatically contains $WX$ so you just need to count the permutations that contain $WX$, no?
– lulu
Nov 16 at 15:39
but then it won't have the V always, right ? then you can't count it for the 1st condition
– Nour M Al-S
Nov 16 at 15:48
I don't understand. You said "or", so I only need to pass one of the tests. Or did you mean to use the "Exclusive Or", in which case you are asking for those permutations which contain $WX$ but which do not contain $VWX$, If the latter, you really need to clarify that in your post...in standard English "or" is non-exclusive.
– lulu
Nov 16 at 15:50
I don't know that's how the question came exactly
– Nour M Al-S
Nov 16 at 15:54
Well, it's your problem. My guess is what I wrote in my first comment...that is, I'd assume that the non-exclusive or was intended. Maybe solve both interpretations! Good practice. Using the non-exclusive or, you are just looking for the permutations that contain $WX$. Count those. Then assume that the non-exclusive or was intended, so you have to subtract off those that contain $VWX$. Neither computation is difficult.
– lulu
Nov 16 at 15:58
|
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I know how to calculate each task like the following
7! + 8!
but I am not sure how to subtract the overlapping part
usually, it will the (or) parts is between distinct string, but in this example, they share 2 letters
how do I solve this?
discrete-mathematics permutations
asked Nov 16 at 15:36
Nour M Al-S
1
I know how to calculate each task like the following
7! + 8!
but I am not sure how to subtract the overlapping part
usually, it will the (or) parts is between distinct string, but in this example, they share 2 letters
how do I solve this?
discrete-mathematics permutations
discrete-mathematics permutations
asked Nov 16 at 15:36
Nour M Al-S
1
asked Nov 16 at 15:36
Nour M Al-S
1
asked Nov 16 at 15:36
Nour M Al-S
1
asked Nov 16 at 15:36
Nour M Al-S
1
asked Nov 16 at 15:36
Nour M Al-S
1
1
3
If the permutation contains $VWX$ then it automatically contains $WX$ so you just need to count the permutations that contain $WX$, no?
– lulu
Nov 16 at 15:39
but then it won't have the V always, right ? then you can't count it for the 1st condition
– Nour M Al-S
Nov 16 at 15:48
I don't understand. You said "or", so I only need to pass one of the tests. Or did you mean to use the "Exclusive Or", in which case you are asking for those permutations which contain $WX$ but which do not contain $VWX$, If the latter, you really need to clarify that in your post...in standard English "or" is non-exclusive.
– lulu
Nov 16 at 15:50
I don't know that's how the question came exactly
– Nour M Al-S
Nov 16 at 15:54
Well, it's your problem. My guess is what I wrote in my first comment...that is, I'd assume that the non-exclusive or was intended. Maybe solve both interpretations! Good practice. Using the non-exclusive or, you are just looking for the permutations that contain $WX$. Count those. Then assume that the non-exclusive or was intended, so you have to subtract off those that contain $VWX$. Neither computation is difficult.
– lulu
Nov 16 at 15:58
|
show 1 more comment
3
If the permutation contains $VWX$ then it automatically contains $WX$ so you just need to count the permutations that contain $WX$, no?
– lulu
Nov 16 at 15:39
but then it won't have the V always, right ? then you can't count it for the 1st condition
– Nour M Al-S
Nov 16 at 15:48
I don't understand. You said "or", so I only need to pass one of the tests. Or did you mean to use the "Exclusive Or", in which case you are asking for those permutations which contain $WX$ but which do not contain $VWX$, If the latter, you really need to clarify that in your post...in standard English "or" is non-exclusive.
– lulu
Nov 16 at 15:50
I don't know that's how the question came exactly
– Nour M Al-S
Nov 16 at 15:54
Well, it's your problem. My guess is what I wrote in my first comment...that is, I'd assume that the non-exclusive or was intended. Maybe solve both interpretations! Good practice. Using the non-exclusive or, you are just looking for the permutations that contain $WX$. Count those. Then assume that the non-exclusive or was intended, so you have to subtract off those that contain $VWX$. Neither computation is difficult.
– lulu
Nov 16 at 15:58
3
3
If the permutation contains $VWX$ then it automatically contains $WX$ so you just need to count the permutations that contain $WX$, no?
– lulu
Nov 16 at 15:39
If the permutation contains $VWX$ then it automatically contains $WX$ so you just need to count the permutations that contain $WX$, no?
– lulu
Nov 16 at 15:39
but then it won't have the V always, right ? then you can't count it for the 1st condition
– Nour M Al-S
Nov 16 at 15:48
but then it won't have the V always, right ? then you can't count it for the 1st condition
– Nour M Al-S
Nov 16 at 15:48
I don't understand. You said "or", so I only need to pass one of the tests. Or did you mean to use the "Exclusive Or", in which case you are asking for those permutations which contain $WX$ but which do not contain $VWX$, If the latter, you really need to clarify that in your post...in standard English "or" is non-exclusive.
– lulu
Nov 16 at 15:50
I don't understand. You said "or", so I only need to pass one of the tests. Or did you mean to use the "Exclusive Or", in which case you are asking for those permutations which contain $WX$ but which do not contain $VWX$, If the latter, you really need to clarify that in your post...in standard English "or" is non-exclusive.
– lulu
Nov 16 at 15:50
I don't know that's how the question came exactly
– Nour M Al-S
Nov 16 at 15:54
I don't know that's how the question came exactly
– Nour M Al-S
Nov 16 at 15:54
Well, it's your problem. My guess is what I wrote in my first comment...that is, I'd assume that the non-exclusive or was intended. Maybe solve both interpretations! Good practice. Using the non-exclusive or, you are just looking for the permutations that contain $WX$. Count those. Then assume that the non-exclusive or was intended, so you have to subtract off those that contain $VWX$. Neither computation is difficult.
– lulu
Nov 16 at 15:58
Well, it's your problem. My guess is what I wrote in my first comment...that is, I'd assume that the non-exclusive or was intended. Maybe solve both interpretations! Good practice. Using the non-exclusive or, you are just looking for the permutations that contain $WX$. Count those. Then assume that the non-exclusive or was intended, so you have to subtract off those that contain $VWX$. Neither computation is difficult.
– lulu
Nov 16 at 15:58
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
If repetition is not allowed and order isn't important, then you have $6!$ ways to do the string.
But note that, if you have the events
$E_1$: The string contains WX
&
$E_2$: The string contains VWX
We have that $E_1 cup E_2= E_1$
So, I think, it is enough to say in your title that '... contain the string WX'
answered Nov 16 at 15:54
Fareed AF
36711
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If repetition is not allowed and order isn't important, then you have $6!$ ways to do the string.
But note that, if you have the events
$E_1$: The string contains WX
&
$E_2$: The string contains VWX
We have that $E_1 cup E_2= E_1$
So, I think, it is enough to say in your title that '... contain the string WX'
answered Nov 16 at 15:54
Fareed AF
36711
add a comment |
up vote
0
down vote
If repetition is not allowed and order isn't important, then you have $6!$ ways to do the string.
But note that, if you have the events
$E_1$: The string contains WX
&
$E_2$: The string contains VWX
We have that $E_1 cup E_2= E_1$
So, I think, it is enough to say in your title that '... contain the string WX'
answered Nov 16 at 15:54
Fareed AF
36711
add a comment |
up vote
0
down vote
up vote
0
down vote
If repetition is not allowed and order isn't important, then you have $6!$ ways to do the string.
But note that, if you have the events
$E_1$: The string contains WX
&
$E_2$: The string contains VWX
We have that $E_1 cup E_2= E_1$
So, I think, it is enough to say in your title that '... contain the string WX'
answered Nov 16 at 15:54
Fareed AF
36711
If repetition is not allowed and order isn't important, then you have $6!$ ways to do the string.
But note that, if you have the events
$E_1$: The string contains WX
&
$E_2$: The string contains VWX
We have that $E_1 cup E_2= E_1$
So, I think, it is enough to say in your title that '... contain the string WX'
answered Nov 16 at 15:54
Fareed AF
36711
answered Nov 16 at 15:54
Fareed AF
36711
answered Nov 16 at 15:54
Fareed AF
36711
answered Nov 16 at 15:54
Fareed AF
36711
36711
add a comment |
add a comment |
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3
If the permutation contains $VWX$ then it automatically contains $WX$ so you just need to count the permutations that contain $WX$, no?
– lulu
Nov 16 at 15:39
but then it won't have the V always, right ? then you can't count it for the 1st condition
– Nour M Al-S
Nov 16 at 15:48
I don't understand. You said "or", so I only need to pass one of the tests. Or did you mean to use the "Exclusive Or", in which case you are asking for those permutations which contain $WX$ but which do not contain $VWX$, If the latter, you really need to clarify that in your post...in standard English "or" is non-exclusive.
– lulu
Nov 16 at 15:50
I don't know that's how the question came exactly
– Nour M Al-S
Nov 16 at 15:54
Well, it's your problem. My guess is what I wrote in my first comment...that is, I'd assume that the non-exclusive or was intended. Maybe solve both interpretations! Good practice. Using the non-exclusive or, you are just looking for the permutations that contain $WX$. Count those. Then assume that the non-exclusive or was intended, so you have to subtract off those that contain $VWX$. Neither computation is difficult.
– lulu
Nov 16 at 15:58