Question 1164376: This is more like math-language problem of mine rather than actual math.
The question starts with "In how many ways n women be seated in a row so that 2 PARTICULAR women will not be next to each other?" Here, my understanding is that those 2 women can be treated as one entity and order does not matter ( Whether they sit Lt and Rt or Rt and Lt), and there is no need to invoke 2! ( I know that 2! is 2 and it does not matter here, but had it been 3 or more then whether it is expressed as 3! or 3 will matter, as you know).
Then it goes on, as part of the solution, "Wih no restrictions, n women may be seated in a row in nPn ways. I f 2 of the n women must always sit next to each other, the number of arrangements = 2!(n-1Pn-1)." Here I think...although I am not sure...2! is appropriate, since there is no "Particular". If there was a "Particular"..2! will not be appropriate and only require 2. (Again please ref above regarding my comment about had it been >2)
And then the solutions goes on to, "Hence the number of ways n women can be seated in a row if 2 PARTICULAR women may never sit together = nPn - 2(n-1Pn-1)." ...2 being the factorial of 2 in question.
What happened it being "Particular" to start with, and how does substracting ( yes... once in a way I can't help using Chauser English for subtraction) arrangements of any two women equates (mathematically!) any two Particular women?
Please dont tell me that I am hung up too much on those two PARTICULAR women !! (=
Just tell me where I am getting turned around. I really need to understand this and get going so I dont insert or not insert n! inappropriately!
Please help
Thanks
Answer by ikleyn(52756)   (Show Source):
You can put this solution on YOUR website! .
Solution
In all, there are n! different possible orderings/arrangements/permutations of "n" women in the row.
From this number, we should subtract the number of those permutations, where two women are seat together.
The number of such permutations is 2*(n-1)! (we threat this particular pair of women as one object and permutate then (n-1) objects;
the multiplier 2 is to account for two different permutations of the type (Alicia,Barbara) and (Barbara,Alicia) inside this particular pair.
So, the final answer is N = n! - 2*(n-1)!.
-------------
Solved.
It is a standard method solving this problem, and it is very close to your logic.
================
Comment from student:
I understand that fully well. But my problem is the second paragraph: the number of arrangements = 2!(n-1Pn-1)."
Here I think...although I am not sure...2! is appropriate, since there is no "Particular".
If there was a "Particular"..2! will not be appropriate and only require 2.
(Again please ref above regarding my comment about had it been >2) So is the 2! is a typo in the book?
I am a beginner. I am not willing to call anything in a math book a typo, until I check with a teacher;
lest, >70% of answers will be typos for me!!(= Thank you
My response :
You may treat the number of interior permutations of two particular women as 2! or 2.
Since BOTH these values are the same, there is no "error" or "typo" in this case.
Did I fully answer your question ?
/\/\/\/\/\/\/\
Comment from student :
Not really. It is ok that 2!=2 here and the answer did not change. My question is what if it was 3 PARTICULAR woman.
Then why is it 3! since they are being treated as one entity? Obviously the answers are going to be different with 3 Vs 3!
My response :
With 3 particular women the answer is n! - 3!*(n-2)! = n! - 6*(n-2)!
And it is different problem then.
So, I do not really understand your truble.
/\/\/\/\/\/\/\/
/\/\/\/\/\/\/\/
Comment from student: When we are treating the 3 women as one entity when the order does not matter how they sit,
why 3! (that leads to 6) ...why not just 3? 3! means that the 3 women can be seated in so (6) many different ways,
which is not relevent if you are going to treat them sitting together as one entity...right ?
My response : You ask me " . . . right ?".
My response is "WRONG".
We consider these three woman as one object from the "remoted point of view";
but we MUST ACCOUNT that 3! = 6 interior permutations are possible inside this entity/object.
Honestly, I just tired to explain these simple things.
A prepared students must "catch it" from the air.
Have a nice day (!)
\\\\\\\\\\\\\\\\\
Yours misunderstanding starts from the point that you NEGLECT to read
my wording solution from the beginning to the end.
If you want to get understanding, you MUST read and RE-READ it until
understanding will COME to your mind suddenly and in full.
I just did my part. Now YOU should do yours.
/\/\/\/\/\/\/\/\/
/\/\/\/\/\/\/\/\/
/\/\/\/\/\/\/\/\/
Comment from student : If the answer to my question was worded: "The 2! is not there because they are sitting together,
where 2 will suffice. It is there because the final answer demands not when they are sitting together,
but when they are NOT sitting together."
These two sentences would have put my question to rest, without needing any didactic explanations.!
But never mind. I was thinking about it all night and I got it on my own. Now it will be etched for ever in my mind,
essentially due to our tedious conversations! So all is not a waste, afterall...at leat for me!
My response : My congrats (! ! !) Finally (!)
Exactly as I said "If you want to get understanding, you MUST read and RE-READ it until
understanding will COME to your mind suddenly and in full."
I am very glad, at the end.
Next time, when you want to have two-way conversation, please refer to the problem's ID number.
I just explained to you once, what it is and where to find it, but I am not sure if you absorbed my explanations.
|
|
|
