Question 1207318: Hi can you help me with this problem. Does it matter what the couple are, or do
we just say two people are a couple.
The drama club buys tickets for 12 seats in a row for a local production, and
then sends the tickets randomly to the 12 people who ordered seats. Of these 12
people, 7 are female and 5 are male. Determine the probability that:
a) A couple will receive tickets sitting together.
b) The women will be seated together and the men will be seated together.
Thank you
Found 3 solutions by ikleyn, Edwin McCravy, math_tutor2020:
Answer by ikleyn(52896)   (Show Source):
You can put this solution on YOUR website! .
Hi can you help me with this problem. Does it matter what the couple are, or do we just say two people are a couple.
The drama club buys tickets for 12 seats in a row for a local production, and then sends the tickets randomly
to the 12 people who ordered seats. Of these 12 people, 7 are female and 5 are male. Determine the probability that:
a) A couple will receive tickets sitting together.
b) The women will be seated together and the men will be seated together.
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Obviously, your question relates to the word " couple " in question (a).
This notion/conception/object " couple " is NOT defined in the body of the problem, neither explicitly nor implicitly;
therefore, formally, there is no proper logical connection between the body of the problem and the question.
It is very BAD (inappropriate) style writing/formulating/presenting Math problems.
I would say, that due to this deficiency, the given text is not a Math problem, at all.
Simply soup of words, instead.
From the post, it is clear that a person who created this composition,
has no right idea what is a Math problem, at all.
I see such defective compositions every day at this forum; sometimes, even several times per day.
Simply many people, who think that they create Math problems and disseminate their writing
in the Internet, do not have necessary professional skills to write Math properly.
Of course, if reading soup of words is fun for you, then you won't care about these details.
But if you come for knowledge and want to learn to speak mathematically in a right way,
then you must pay the closest attention to the accuracy of your language.
There are no math problems without internal harmony in their formulations.
Answer by Edwin McCravy(20064)  (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answers:
(a) 1/6
(b) 1/396
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Explanation for Part (a)
The first 12 letters of the English alphabet are A through L.
Let A and B represent the two people that make up this particular couple that want to sit together.
Let X represent their place in any given permutation.
The sequence C,D,X,E,F,G,H,I,J,K,L is one such example.
We'll replace X with either A,B or B,A.
Doing it this way guarantees that A & B stick together.
i.e. it guarantees that A & B are next door neighbors.
There are initially 12 letters.
Removing A & B temporarily drops things to 12-2 = 10 letters.
Temporarily introducing X bumps it up to 10+1 = 11 letters.
There are 2*11! ways to arrange the 12 people so that A,B sit together in either order.
The exclamation mark indicates factorial.
The 11! factorial term is the number of ways to arrange the 11 letters involving X (but not A,B just yet), and the 2 out front is to consider the two cases when A,B happens or B,A happens.
This is out of 12! ways to arrange all 12 people regardless if that particular couple is sitting together or not.
So we have:
2*11! = number of ways to arrange people so A,B stick together
12! = number of permutations total
Divide the two items mentioned:
(2*11!)/(12!)
(2*11!)/(12*11!)
2/12
1/6 is the answer to part (a).
Notice the 11! terms cancel which is really convenient.
Another thing to note is if you had n people then 2/n is the probability that one particular couple sticks together.
The proof is short so I'll leave it to the reader.
Hint: 2*(n-1)! is the number of ways to arrange n people so that exactly one couple sticks together.
I recommend trying small values of n to see examples of why this formula works.
Here's a small bit of scratch work when n = 3
The double star notation indicates when A,B are together
ABC **
ACB
BAC **
BCA
CAB **
CBA **
There are 4 cases when A,B are together out of 6 permutations.
4/6 = 2/3 is the probability of A,B being together in either order.
This matches with the formula 2/n where n = 3. So this example helps confirm the formula works. I recommend trying it with n = 4 to see what you find.
A similar question is found here
https://math.stackexchange.com/questions/114367/probability-people-sitting-in-a-row-linear-arrangement
The "solution" 1/63 mentioned on that page is NOT correct. I'm not sure why the professor made such a strange error. Instead the solution is 1/5 as the other people point out.
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Explanation for Part (b)
Let M and W represent the block of men and block of women respectively.
We could have MW or WM
So there are 2 ways to arrange these blocks.
Within any given such arrangement, we have 7! ways to arrange the 7 women within their respective block, and 5! ways to arrange the 5 men in their respective block.
2*7!*5! is the number of ways to arrange the people so that the genders are fully separated.
This is out of 12! ways to arrange all 12 people regardless if the genders are blocked together or not.
Divide the items mentioned.
(2*7!*5!)/(12!)
(2*7!*5!)/(12*11*10*9*8*7!)
(2*5!)/(12*11*10*9*8)
(2*5*4*3*2*1)/(12*11*10*9*8)
(4*3*2*1)/(6*11*2*9*8)
(2*1)/(6*11*2*3*2)
(1)/(6*11*2*3)
1/(66*6)
1/396 is the answer to part (b)
The steps above hopefully show how I cancelled things. Other routes are possible.
You can use a calculator to make quick work of this, but it still might be useful to know how to do this by hand.
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