How do I approach b and c in this question(combinatorics)?

Atrox

Atrox

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As far as I know, I have to discount from all variants (which is 1001) the number of variants when the 2 people are together for b). However, I do not understand how to select these 2 people being together, do I take them as 1?
 
I got the same answer. Good job.

If you show us how you got your answer we might be able to show you a clearer way of thinking about it.
 
Steven G said:
I got the same answer. Good job.

If you show us how you got your answer we might be able to show you a clearer way of thinking about it.
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these questions have the answers, but I do not know how to solve them that is why I am asking for help.
 
solitude said:
View attachment 33158

As far as I know, I have to discount from all variants (which is 1001) the number of variants when the 2 people are together for b). However, I do not understand how to select these 2 people being together, do I take them as 1?
Click to expand...
What we'd like to see, in order to help you most effectively, is a demonstration of the actual work you have in mind when you say "take them as 1". That describes a valid method of solution. If you show an actual attempt, rather than just a description, we can tell you whether you are doing the right thing, or need a little correction. In fact, making an attempt will commonly raise more questions in your mind, which you can either answer for yourself, or ask us!
 
solitude said:
As far as I know, I have to discount from all variants (which is 1001) the number of variants when the 2 people are together for b). However, I do not understand how to select these 2 people being together, do I take them as 1?
Click to expand...
The answer for b would be...
1001 - (the number of ways that the two specific candidates can be sent together)

Imagine that these two candidates HAVE been "pre-assigned" - now how many ways can the remaining places be allocated?
 
Part C
Do this in two parts
1st part: Do not include the friends
2nd part: Do include the friends.
 
Cubist said:
The answer for b would be...
1001 - (the number of ways that the two specific candidates can be sent together)

Imagine that these two candidates HAVE been "pre-assigned" - now how many ways can the remaining places be allocated?
Click to expand...

There's another way of solving part b using addition only. Let's label the awkward candidates. Alex and Bernadette don't want to be sent together. Now you can add together the following three scenarios...
  • Alex doesn't attend AND Bernadette doesn't attend, so how many combinations of 4 are now possible?
  • Alex attends (therefore Bernadette doesn't), so how many ways can the other 3 places be filled?
  • ???
This isn't better or worse than the other method, but it might help to improve your knowledge.
 
Dr.Peterson said:
What we'd like to see, in order to help you most effectively, is a demonstration of the actual work you have in mind when you say "take them as 1". That describes a valid method of solution. If you show an actual attempt, rather than just a description, we can tell you whether you are doing the right thing, or need a little correction. In fact, making an attempt will commonly raise more questions in your mind, which you can either answer for yourself, or ask us!
Click to expand...
I don't know how to select the 2 candidates together. but I would take it as 1 candidate which would result in C(13,4). So, C(14,4)-C(13,4)=1001-715=286
 
Cubist said:
The answer for b would be...
1001 - (the number of ways that the two specific candidates can be sent together)

Imagine that these two candidates HAVE been "pre-assigned" - now how many ways can the remaining places be allocated?
Click to expand...
I see that too, but I do not know how to calculate (the number of ways that the two specific candidates can be sent together)
 
solitude said:
I don't know how to select the 2 candidates together. but I would take it as 1 candidate which would result in C(13,4). So, C(14,4)-C(13,4)=1001-715=286
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This is for (b), right?

You presumably want to subtract the number of ways the two could both be included. To do that, you are imagining them being tied together and treated as one person. But then they would both have to fit in one space, so to speak!

A key to solving this sort of problem is to hold off on writing numbers, and give lots of time to thinking about what you are doing. Rather than just write C(13,4), you need to imagine what those 13 people are, and the 4 that are selected, and see why that doesn't accomplish the goal.

Since choosing two people is not the same as choosing one, you have to treat them separately: if they are both chosen, then how many are left for the remaining places, and how many such places are there? See #6.
 
solitude said:
I see that too, but I do not know how to calculate (the number of ways that the two specific candidates can be sent together)
Click to expand...
Cubist said:
Imagine that these two candidates HAVE been "pre-assigned" - now how many ways can the remaining places be allocated?
Click to expand...

OK, Sheila and Fred are BOTH definitely going. Therefore, how many of the 14 possible candidates remain (since both Sheila and Fred are in the list of 14)? And how many more can be sent (how many places are left)?
 
Cubist said:
OK, Sheila and Fred are BOTH definitely going. Therefore, how many of the 14 possible candidates remain (since both Sheila and Fred are in the list of 14)? And how many more can be sent (how many places are left)?
Click to expand...
Okay so it is going to be C(12,2) since they are selected for sure therefore we are selecting from 4-2 and 14-2, is that correct?
 
Following Dr. Peterson’s advice, think about this.

A and B will not go together.

We know the total number of delegations is 1001.

How many delegations can we form that include both A and B? [imath]\dbinom{2}{2} * \dbinom{12}{2}.[/imath]

The number of delegations that do not include both A and B must be the total less those that do include both A and B.

[math]1001 - \dbinom{2}{2} * \dbinom{12}{2} = 1001 - 1 * \dfrac{12 * 11}{2} = 1001 - 66 = 935.[/math]
Dr Peterson frequently recommends thinking about combinatorial problems a second way as a check on our reasoning.

How many delegeations can we form if we include neither A nor B? [imath]\dbinom{2}{0} * \dbinom{12}{4}.[/imath]

How many delegations can we form if we include person A but not B? [imath]\dbinom{1}{1} * \dbinom{1}{0} * \dbinom{12}{3}.[/imath]

How many delegations can we form if we include person B but not A? [imath]\dbinom{1}{1} * \dbinom{1}{0} * \dbinom{13}{3}.[/imath]

Are those sets mutually exclusive? Yes.

[math]\dbinom{2}{0} * \dbinom{12}{4} + \dbinom{1}{1} * \dbinom{1}{0} * \dbinom{12}{3} + \dbinom{1}{1} * \dbinom{1}{0} * \dbinom{13}{3} =\\ 1 * 1 * \dbinom{12}{4} + 2 * 1 * 1 * \dbinom{12}{3} = \dfrac{12 * 11 * 10 * 9}{4 * 3 * 2} + 2 * \dfrac{12 * 11 * 10}{3 * 2} = \\ 99 * 5 + 2 * 2 * 11 * 10 = 495 + 440 = 935 \ \checkmark [/math]
Think carefully before doing computations.

Try to confirm your answer by a different method.
 
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solitude said:
View attachment 33158

As far as I know, I have to discount from all variants (which is 1001) the number of variants when the 2 people are together for b). However, I do not understand how to select these 2 people being together, do I take them as 1?
Click to expand...
Part C
Include both friends: So we need to choose 2 more from the 12 remaining people (14-2=12)
Don't include both friends: So we need to choose 4 from 12.
Add these together.
 
Dr Peterson is correct when he said that you need to think about this problem.
I personally wrote out the numbers 1 through 14. I circled 1 and 2 as friends.
Then, at least for me, the problem became simple.

There will be a simple way for you to see it and it can surely be different then how others see it.
I am trained in math and another helper here is trained in computer science. We always do probability problems differently but (usually) in clear way. You will develop your own style, but this is not to say that you can't learn from our technique.

Dr Peterson and my probability teachers drilled into my brain by repeatedly saying that you should do a probability problem two different ways and if you get the same answer then you are probably correct. That is, you really should know how to do these problems at least two ways.
 
Steven G said:
Dr Peterson and my probability teachers drilled into my brain by repeatedly saying that you should do a probability problem two different ways and if you get the same answer then you are probably correct. That is, you really should know how to do these problems at least two ways.
Click to expand...
Great advice :thumbup:
 
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