p&c q12

[Saumyojit]

A is a set containing 'n' elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of P. A subset Q of A is chosen again at random. Find the no of ways choosing p &q so that P intersection Q contains exactly 1 element.​

All the subsets that are possible out of n elements = 2^(n) -1
No of ways P is chosen = 2^(n)-1 C1
Cardinality of |P|=no
No of ways Q is chosen = 2^(no)-1 C1 =

Then how to proceed

@Dr.Peterson

 

[Dr.Peterson]

A is a set containing 'n' elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of P. A subset Q of A is chosen again at random. Find the no of ways choosing p &q so that P intersection Q contains exactly 1 element.​

All the subsets that are possible out of n elements = 2^(n) -1
No of ways P is chosen = 2^(n)-1 C1
Cardinality of |P|=no
No of ways Q is chosen = 2^(no)-1 C1 =

Then how to proceed

I wouldn't start that way, so I can't tell you how to proceed from there.

For one thing, P could be all of A, or it could be empty (ignoring for the moment the requirement that the intersection is nonempty), so your count is wrong, if it matters. Second, I have no idea what you are doing with Q. Are you thinking that is a subset of P, rather than of A?

The first thing to do, as always, is to be sure what the problem says. The reconstruction idea seems confusing; I think all it's saying is something like this: If you pick a subset P of A (not removing them, just choosing), and then I independently choose a subset Q of A, and it turns out that the intersection of P and Q contains exactly one element, how many ways could that happen?

I'd start by picturing a Venn diagram, and counting how many ways there are to fill it in (that is, placing the n elements of A into the diagram). If nothing else, that makes it easier to picture what is being done. But I think it leads to a very simple answer.

 

[Saumyojit]

If you pick a subset P of A (not removing them, just choosing), and then I independently choose a subset Q of A, and it turns out that the intersection of P and Q contains exactly one element, how many ways could that happen?

didn't understand

 

[Dr.Peterson]

didn't understand

I can't tell what you don't understand. Please show your interpretation of what you think I might have said, so we have more to discuss.

Do you not understand that both P and Q are subsets of A, rather than Q being a subset of P?

Did you try making a Venn diagram as I suggested? Maybe try solving the problem for a small value of n, such as 2 or 3, by drawing all possible Venn diagrams that fit the description, and show them to us, so we can see what you are missing.

 

[Saumyojit]

Do you not understand that both P and Q are subsets of A, rather than Q being a subset of P?

The set A is reconstructed by replacing the elements of P means P and Q are subsets of A

if A= 1,2
P=1 or 2 or 1,2
After reconstruction of A=what elements will A have?

If A remain same then,
Q can be 1 or 2 or 1,2 --> I independently chose a subset Q of new A
or
A now consists of only P element/s .If p is 1 then new A contains one?

What does replacing the elements of P means then?

 

[Dr.Peterson]

The set A is reconstructed by replacing the elements of P means P and Q are subsets of A

if A= 1,2
P=1 or 2 or 1,2
After reconstruction of A=what elements will A have?

If A remain same then,
Q can be 1 or 2 or 1,2 --> I independently chose a subset Q of new A
or
A now consists of only P element/s .If p is 1 then new A contains one?

What does replacing the elements of P means then?

As I said, the wording is confusing. I have no idea why they chose to say what they did, unless perhaps in their context they had been doing problems where "choose" means "remove". (It does not.)

As I understand it, "The set A is reconstructed by replacing the elements of P" means simply that, having removed some elements of set A to form set P, they put them back, so that set A is just what it was to begin with. Nothing else makes sense; if any elements in A were changed, it would no longer be A.

Once again, I have to ask you: Where in the world did this question come from??

Here is a definition of "replace"; your problem uses the fourth meaning: https://www.dictionary.com/browse/replace

 

[Saumyojit]

Where in the world did this question come from??

https://www.toppr.com/ask/question/a-is-a-set-containing-n-elements-a-subset-p-of-a-is-chosen-at-6/

a similar type

​

​

Now,
if A= 1,2 ---> P=1 or 2 or 1,2
Q can be 1 or 2 or 1,2
right?

 

[Dr.Peterson]

Now,
if A= 1,2 ---> P=1 or 2 or 1,2
Q can be 1 or 2 or 1,2
right?

True, but that doesn't answer the question! Not all pairs P and Q work.

How many ways are there? I strongly recommend drawing the Venn diagrams, as I suggested, because that is how I saw the solution to the problem (though I only had to imagine them).

I had seen this and several other pages when I searched for the troublesome sentence. But note that I asked not where you found it, but where it came from -- that is, who wrote such a poorly written problem in the first place, because it needs context. Unfortunately, sites such as this rarely provide such contextual information. That's among many reasons I would not try to learn from such sites. (In addition to the fact that the people who write the answers are surely paid even less than the poor drudges who write answers for textbooks ...)

 

[Saumyojit]

drawing the Venn diagrams

if A= 1,2 ---> P=1 or 2 or 1,2
Q can be 1 or 2 or 1,2

when p=1, Q can be 1 or 2 or 1,2 . so 2 ways of choosing p &q exactly one element in common .


when p=2, Q can be 1 or 2 or 1,2 . so 2 ways of choosing p &q exactly one element in common .

when p=1,2 Q can be 1 or 2 or 1,2 . so 2 ways of choosing p &q exactly one element in common .

6 ways when n=2.

So, n* 3^(n-1) which is the correct answer .

this is i have done by taking a small no n and by breaking down .


is this the right method ?



Replace means to restore to a former or the proper place:
thanks for this

 

[Dr.Peterson]

6 ways when n=2.

So, n* 3^(n-1) which is the correct answer .

That's the correct answer. But do you see how to derive that formula? One example isn't enough.

is this the right method ?

I see no method yet; but there are probably several valid methods, so don't ask for "the right one".

How can you fill in the Venn diagram? First pick an item to put in the intersection, then ...

 

[Cubist]

I'd start by picturing a Venn diagram

Did you try making a Venn diagram as I suggested?

I strongly recommend drawing the Venn diagrams

I don't want @Dr.Peterson 's keyboard to wear out

 

[Saumyojit]

I don't want @Dr.Peterson 's keyboard to wear out

View attachment 28351

First i have to choose in how many ways can I choose that one element out of n elements which is common --> nC1
then , There are n-1 elements in A . Each of these n-1 elements can go to either subsets p or q or nowhere in one particular event. 1st element has three choices , 2nd element has three choices , 3rd element has three options ....
nc1 * 3^(n-1)


this way of thinking didn't came to my mind.

 

[Cubist]

First i have to choose in how many ways can I choose that one element out of n elements which is common --> nC1
then , There are n-1 elements in A . Each of these n-1 elements can go to either subsets p or q or nowhere in one particular event. 1st element has three choices , 2nd element has three choices , 3rd element has three options ....
nc1 * 3^(n-1)

Excellent!

this way of thinking didn't came to my mind.

You've got it now. Just keep practicing, that's the way to gain experience. Hopefully one day you'll be able to imagine the diagram in your head like @Dr.Peterson