How many permutations of the letters ABCDEFG contain the string BCD?

[Havie]

Hello all,
I have come back to college after a # of years out of school,
I am struggling with a few concepts in my discrete math class and some further help would be greatly appreciated as I have exhausted all other tutoring avenues.

Question:
How many permutations of the letters ABCDEFG contain:

a) The String BCD?

At first I didnt understand that the order of BCD mattered, for example CDB does not count. So I went about solving this the total wrong way,
Question 1) what part of this problem is clarifying that rule? I feel like it could go either way and is almost up for interpretation .
Question 2)
So i eventually figured out that since order matters you group BCD as 1 item
so its
BCD X A X E X F X G
which is
5!

what I dont understand is exactly the distribution of how this is playing out.

For example,
I would think
BCD X A = ABCD or ABCD (2!)
then
BCD X A X E = AEBCD, ABCDE, EABCD, EBCDA, BCDAE, BCDEA which is (3!)

makes sense, but what about:
EBCD, BCDE

what is handeling this portion of the options?

Thanks for your help and time.

 

[Dr.Peterson]

Hello all,
Question:
How many permutations of the letters ABCDEFG contain:

a) The String BCD?

At first I didnt understand that the order of BCD mattered, for example CDB does not count. So I went about solving this the total wrong way,
Question 1) what part of this problem is clarifying that rule? I feel like it could go either way and is almost up for interpretation .

The word "string" refers to an ordered list (without reference to repetitions). It isn't used as often in math as, say, "permutation", so you can be excused for not being familiar with it, but it is defined. And it surely means the same thing outside of math.

Question 2)
So i eventually figured out that since order matters you group BCD as 1 item
so its
BCD X A X E X F X G
which is
5!

what I dont understand is exactly the distribution of how this is playing out.

For example,
I would think
BCD X A = ABCD or ABCD (2!)
then
BCD X A X E = AEBCD, ABCDE, EABCD, EBCDA, BCDAE, BCDEA which is (3!)

makes sense, but what about:
EBCD, BCDE

what is handeling this portion of the options?

I'm not sure what you mean by "X" in your notation. Can you explain? Is this something you were taught, or your own invention?

The idea here is simply that you treat "BCD" as a single item, so that you are permuting five things, [BCD], [A], [E], [F], and [G]. There are 5! ways to do this. One way to explain it is that you first pick one of the five (5 ways to choose), then a second (4 left to choose from), and so on: 5*4*3*2*1.

If you are still unsure, please explain more fully what you think is not being handled properly.

 

[Havie]

The word "string" refers to an ordered list (without reference to repetitions). It isn't used as often in math as, say, "permutation", so you can be excused for not being familiar with it, but it is defined. And it surely means the same thing outside of math.



I'm not sure what you mean by "X" in your notation. Can you explain? Is this something you were taught, or your own invention?

The idea here is simply that you treat "BCD" as a single item, so that you are permuting five things, [BCD], [A], [E], [F], and [G]. There are 5! ways to do this. One way to explain it is that you first pick one of the five (5 ways to choose), then a second (4 left to choose from), and so on: 5*4*3*2*1.

If you are still unsure, please explain more fully what you think is not being handled properly.

hello, thanks for the reply. the string definition was helpful.

the X meant multiplication sorry.

what I am getting at is :

So i eventually figured out that since order matters you group BCD as 1 item
so its
BCD * A * E * F * G
which is
5!

what I dont understand is exactly the distribution of how this is playing out.

For example,
I would think
BCD * A = ABCD , ABCD (2!)
then
BCD * A * E = AEBCD, ABCDE, EABCD, EBCDA, BCDAE, BCDEA which is (3!)

makes sense, but what about:
EBCD, BCDE

what is handeling this portion of the options?

 

[Dr.Peterson]

hello, thanks for the reply. the string definition was helpful.

the X meant multiplication sorry.

what I am getting at is :

So i eventually figured out that since order matters you group BCD as 1 item
so its
BCD * A * E * F * G
which is
5!

what I dont understand is exactly the distribution of how this is playing out.

For example,
I would think
BCD * A = ABCD , ABCD (2!)
then
BCD * A * E = AEBCD, ABCDE, EABCD, EBCDA, BCDAE, BCDEA which is (3!)

makes sense, but what about:
EBCD, BCDE

what is handeling this portion of the options?

Were you taught to write your calculations as multiplications of things (rather than of numbers) as you are doing? I don't think it's a safe thing to do. BCD isn't a number you can multiply. The notation may be what is confusing you.

You haven't clarified what you mean by this last question, which is what I asked about. EBCD is not among the permutations you are counting, since it has only four letters. So what "potion of the options" do you mean by this? And what do you mean by "handling"?

 

[Havie]

Were you taught to write your calculations as multiplications of things (rather than of numbers) as you are doing? I don't think it's a safe thing to do. BCD isn't a number you can multiply. The notation may be what is confusing you.

You haven't clarified what you mean by this last question, which is what I asked about. EBCD is not among the permutations you are counting, since it has only four letters. So what "potion of the options" do you mean by this? And what do you mean by "handling"?

Hmm,
sorry its hard to explain.

No i wasn't taught any of this, im trying to basically figure it out on my own. before we get into what im asking let me try to explain a little more backstory to my thought.

So you said EBCD is not amoung the permutations? how is it not ? maybe this is where i am confused.
maybe lets start with this:
how many different permutations of ABCDEFG are there? isn't it 7! ?

I know for certain the answer to Question 1) is 5! or 120

so that is saying there is 120 different ways that BCD will occur within the permutations of the string.
one of those permutations will be EBCD, correct?

---------------
so operating under the assumption that is true.
we split our string into
BCD , A , E , F , G
if we assign a # to each of them, thats 5 items
1 , 2 , 3 , 4 , 5

if we multiply them together we get 5! which is = 120 which is the answer.

What I am confused about is basically the general idea behind multiplication and how this is handled. by handled i mean works behind the scenes.

for example
If I wanted to see the different combinations of BCD, A
BCD * A = 2!
BCDA
ABCD

then if i wanted to add in E we are now looking at BCD , A , E
which is = 3!
which means there are 6 different ways to combine these:

AEBCD
ABCDE
EABCD
EBCDA
BCDAE
BCDEA

What confuses me is when I am looking at the bigger picture of 5!
it is 1*2*3*4*5
when you correlate the #s to the letter combinations,
I dont get what is handling ( by handling i mean accounting for) the difference between
BCD*A
and
BCD * E (or any other letter)

I guess im just confused on how multiplication is being applied to distinct objects.




---------------------------
while writing this i got myself even more confused
and heres where that led
----------------------------
if I wanted to see the different combinations of BCD , A , E
It would be 3! , i assume.
So part of that list would be the combinations of just 2 elements in the list and the 3rd element being left out, wouldnt it?
--- maybe im wrong here in including these 2 combinations? but then that destroys the initial argument of different permutations of ABCDEFG , and the string of BCD showing up 120 times.

BCD * A
BCDA
ABCD

BCD * E
BCDE
EBCD

A * E
AE
EA

Then this is where it gets confusing and tricky for me,
problem is theres already 6 items in that list above, and we didnt even handle the combination with the 3rd element.

having to account for those + the 3rd element:


BCD * A
BCDA + E = BCDAE || EBCDA || BCDEA
ABCD + E = ABCDE || EABCD || AEBCD

BCD * E
BCDE + A = BCDEA || ABCDE || BCDAE
EBCD + A = EBCDA || AEBCD || EABCD

A * E
AE + A = AEBCD || BCDAE || ABCDE
EA + A = EABCD || BCDEA || EBCDA


and when you get rid of the redundancies is also 6.

so now theres 6 ( 2 combo) +6 ( 3 combo) = 12

but theres only supposed to 3! which is 6.

So now im just all sorts of lost.
hope some of this comprehensible and you can maybe help catch me on where im going astray in thinking

 

[pka]

Question:
How many permutations of the letters ABCDEFG contain:
a) The String BCD?
At first I didnt understand that the order of BCD mattered, for example CDB does not count. So I went about solving this the total wrong way,
Question 1) what part of this problem is clarifying that rule? I feel like it could go either way and is almost up for interpretation .
Question 2)
So i eventually figured out that since order matters you group BCD as 1 item
so its
BCD X A X E X F X G
which is
5!.

Here is a very useful website for looking up mathematical terms & concepts/
Because you are coming back, if I were you I wound bookmark the site.
As you can see a string is simply a list of individual items which come a certain collection and the items do not have to be unique.
A permutation is a string in which the items are unique. One again that website is a good reference. Click on the permutation.

Given the string ABCDEFG there are 7! different permutations of that string.
Given the string ASEFG there are 5!=120 different permutations of that string. Do you see why?
If the letter S stands for BCD the we know the answer to the original question is 120. Do you see why?

 

[Dr.Peterson]

Hmm,
sorry its hard to explain.

No i wasn't taught any of this, im trying to basically figure it out on my own. before we get into what im asking let me try to explain a little more backstory to my thought.

So you said EBCD is not amoung the permutations? how is it not ? maybe this is where i am confused.
maybe lets start with this:
how many different permutations of ABCDEFG are there? isn't it 7! ?

I know for certain the answer to Question 1) is 5! or 120

so that is saying there is 120 different ways that BCD will occur within the permutations of the string.
one of those permutations will be EBCD, correct?

No, EBCD is not a permutation of ALL of the letters; it is a permutation of only 4 of the 7 letters. The question asks for permutations of the entire string.

So, out of the 7! = 5040 ways to arrange the 7 letters, 5! = 120 contain BCD together, in that order.

---------------
so operating under the assumption that is true.
we split our string into
BCD , A , E , F , G
if we assign a # to each of them, thats 5 items
1 , 2 , 3 , 4 , 5

if we multiply them together we get 5! which is = 120 which is the answer.

This is a simplistic way to think of it. If you try to mechanically equate each item in the string to a number, you will make mistakes in any problem harder than the very simplest.

We are not simply assigning a number to each item; we are counting ways to choose. There are 5 ways to choose the first, 4 ways to choose the second, and so on. So, if anything, I would "assign" the numbers in the opposite order. But they are assigned to choices, not to items.

What I am confused about is basically the general idea behind multiplication and how this is handled. by handled i mean works behind the scenes.

for example
If I wanted to see the different combinations of BCD, A
BCD * A = 2!
BCDA
ABCD

Watch out for words here! These are not "combinations", which is a technical term meaning subsets (without ordering). They are permutations, or orderings.

And the idea is that there are 2 choices for the first, and 1 for the second (because we don't allow repetitions).

then if i wanted to add in E we are now looking at BCD , A , E
which is = 3!
which means there are 6 different ways to combine these:

AEBCD
ABCDE
EABCD
EBCDA
BCDAE
BCDEA

What confuses me is when I am looking at the bigger picture of 5!
it is 1*2*3*4*5
when you correlate the #s to the letter combinations,
I dont get what is handling ( by handling i mean accounting for) the difference between
BCD*A
and
BCD * E (or any other letter)

I guess im just confused on how multiplication is being applied to distinct objects.

There is no difference at all between arranging BCD and A, and arranging BCD and E.

It is becoming even more clear that your faulty model is leading to your confusion.

---------------------------
while writing this i got myself even more confused
and heres where that led
----------------------------
if I wanted to see the different combinations of BCD , A , E
It would be 3! , i assume.
So part of that list would be the combinations of just 2 elements in the list and the 3rd element being left out, wouldnt it?
--- maybe im wrong here in including these 2 combinations? but then that destroys the initial argument of different permutations of ABCDEFG , and the string of BCD showing up 120 times.

BCD * A
BCDA
ABCD

BCD * E
BCDE
EBCD

A * E
AE
EA

Then this is where it gets confusing and tricky for me,
problem is theres already 6 items in that list above, and we didnt even handle the combination with the 3rd element.

You have so far counted the permutations of 3 items "taken 2 at a time", which is written as ₃P₂. These are not part of the permutations of all 3 (since these only include 2), but you could continue from here if you were careful.

having to account for those + the 3rd element:


BCD * A
BCDA + E = BCDAE || EBCDA || BCDEA
ABCD + E = ABCDE || EABCD || AEBCD

BCD * E
BCDE + A = BCDEA || ABCDE || BCDAE
EBCD + A = EBCDA || AEBCD || EABCD

A * E
AE + A = AEBCD || BCDAE || ABCDE
EA + A = EABCD || BCDEA || EBCDA


and when you get rid of the redundancies is also 6.

so now theres 6 ( 2 combo) +6 ( 3 combo) = 12

but theres only supposed to 3! which is 6.

So now im just all sorts of lost.
hope some of this comprehensible and you can maybe help catch me on where im going astray in thinking

At this point, your main error appears to be that you think the permutations you were asked to count don't have to include all items.

Are you taking a course, or reading a book? If so, tell us what it is, so we can have a better idea where you are coming from. If not, you really need to find one, because starting from the beginning, learning what each term means and how each concept is to be thought about, would help a lot.

 

[Havie]

No, EBCD is not a permutation of ALL of the letters; it is a permutation of only 4 of the 7 letters. The question asks for permutations of the entire string.

So, out of the 7! = 5040 ways to arrange the 7 letters, 5! = 120 contain BCD together, in that order.



This is a simplistic way to think of it. If you try to mechanically equate each item in the string to a number, you will make mistakes in any problem harder than the very simplest.

We are not simply assigning a number to each item; we are counting ways to choose. There are 5 ways to choose the first, 4 ways to choose the second, and so on. So, if anything, I would "assign" the numbers in the opposite order. But they are assigned to choices, not to items.



Watch out for words here! These are not "combinations", which is a technical term meaning subsets (without ordering). They are permutations, or orderings.

And the idea is that there are 2 choices for the first, and 1 for the second (because we don't allow repetitions).



There is no difference at all between arranging BCD and A, and arranging BCD and E.

It is becoming even more clear that your faulty model is leading to your confusion.



You have so far counted the permutations of 3 items "taken 2 at a time", which is written as ₃P₂. These are not part of the permutations of all 3 (since these only include 2), but you could continue from here if you were careful.



At this point, your main error appears to be that you think the permutations you were asked to count don't have to include all items.

Are you taking a course, or reading a book? If so, tell us what it is, so we can have a better idea where you are coming from. If not, you really need to find one, because starting from the beginning, learning what each term means and how each concept is to be thought about, would help a lot.

Okay,
Thanks
The idea that were counting different choices not items is what i think i was searching for.
this clears it up alot.

also the permutations of 7 letters being arranged differently have 120 occurances where BCD wind up next to one another WITHIN a 7 length string helps alot as well.

Thanks for clearing that up.

Also I am taking a discrete math in college but the teacher barely explains the core math concepts and just assumes were fluent in all math. The class avg for the first exam was a D. Worst I've ever seen a class struggle. 2nd Exam is Monday and its not looking much better.
Our book is Kenneth Rosen- Discrete mathematics and its applications. which is vaguely helpful. Its quite a challenge to read abstract math concepts condensed into English sentences instead of the usual style where a problem is solved vertically down a sheet of paper.