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Thread: combinations

  1. #1
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    combinations

    There is a hotel.There is 1 room for 2 persons.There are 4 rooms for 3
    persons. 3 Person Rooms are identical.In how many ways 14 people can
    stay in that hotel?


    my work:
    C(14,2).C(12,3).C(9,3).C(6,3)
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  2. #2
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    Re: combinations

    Hey kastamonu.

    Can you explain your answer? This is the most important attribute of mathematics [i.e. the logic and working needed to get the answer - not the answer itself].
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  3. #3
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    Re: combinations

    Quote Originally Posted by kastamonu View Post
    There is a hotel.There is 1 room for 2 persons.There are 4 rooms for 3
    persons. 3 Person Rooms are identical.In how many ways 14 people can
    stay in that hotel?
    @kastamonu, you continue posting the poorest worded questions possible.
    If one answered this question literally as written, then the answer is:
    how many ways can we rearrange $AABBBBBBBBBBBB$ or $\dfrac{14!}{2\cdot 12!}$.
    I do not think the is what you wanted the question to mean. Nor is it your answer.

    However, if you leave the qualifier 3 Person Rooms are identical then it reads:
    how many ways can we rearrange $AABBBCCCDDDEEE$ or $\dfrac{14!}{2\cdot (3!)^4}$.
    That is exactly the answer you gave.

    Which of those two readings do you mean?
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    Re: combinations

    Quote Originally Posted by Plato View Post
    @kastamonu, you continue posting the poorest worded questions possible.
    If one answered this question literally as written, then the answer is:
    how many ways can we rearrange $AABBBBBBBBBBBB$ or $\dfrac{14!}{2\cdot 12!}$.
    I do not think the is what you wanted the question to mean. Nor is it your answer.

    However, if you leave the qualifier 3 Person Rooms are identical then it reads:
    how many ways can we rearrange $AABBBCCCDDDEEE$ or $\dfrac{14!}{2\cdot (3!)^4}$.
    That is exactly the answer you gave.

    Which of those two readings do you mean?
    Pretty sure he means for example that (a,b)(c,d,e)(f,g,h)(i,j,k)(l,m,n) is the same as (a,b)(c,d,e)(i,j,k)(f,g,h)(l,m,n)

    i.e. it's the content of the 3 person rooms that matters not which room it occurs in.

    His formula is correct.
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    Re: combinations

    My solution is wrong.Because the rooms are identical. 14!/2.12! is correct
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    Re: combinations

    I don't understand what did I wrong.What is the difference when rooms are identical?
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    Re: combinations

    Quote Originally Posted by kastamonu View Post
    I don't understand what did I wrong.What is the difference when rooms are identical?
    You did nothing wrong. The way the question is at fault, it is open to multiple readings. The phrase "3 Person Rooms are identical" is the culprit. I would have suggested using the qualifier "With respect to who is rooming with whom, how many ways can fourteen people occupy the five rooms"?

    That kind of is well known in counting and is called an unordered partition.
    Example: How many ways can twelve students be assigned to four study-groups of three each?
    We see that the only aspect that matters is the content of group.
    Now it should be easy to that the groups are identical in purpose but different in content.
    The answer to this is $\dfrac{12!}{(3!)^4\cdot 4!}$

    If we want group $N$ different objects into $0\le k<N$ groups then if $N=qk+r$, the answer is $\dbinom{N}{r}\dfrac{(N-r)!}{(q!)^k(k!)}$
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    Re: combinations

    Maybe C(14,2).2!.12!
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    Re: combinations

    Quote Originally Posted by kastamonu View Post
    Maybe C(14,2).2!.12!
    You are simply guessing. Yo steadfastly refuse to tell us the actual meaning of the question.
    What what does it mean?

    $ABCDEFGHIJKLMN$ is a group of fourteen.
    $(A,L),~[B,C,N],~[D,M,L],~[E,F,K],~\&~[G,H,I]$ two are in a rounded room and the rest assigned to square rooms of three each.
    In this case the 3-persons room are identical except for content.
    There are $\dbinom{14}{2}\cdot\dfrac{12!}{(3!)^4\cdot(4!)}$ different ways to make such an assignment.


    $(A,L),~\underbrace {[B,C,N]}_1,~\underbrace {[D,M,L]}_2,~\underbrace {[E,F,K]}_3~\underbrace {[G,H,I]}_4$
    Now the 3-person rooms are numbered and the answer is $\dbinom{14}{2}\cdot\dfrac{12!}{(3!)^4}$.
    AND that number is actually what you first posted.
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    Re: combinations

    3-person rooms are identical in every way.
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    Re: combinations

    Quote Originally Posted by kastamonu View Post
    3-person rooms are identical in every way.
    Can you read English? Or are you using a translator program?
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    Re: combinations

    I am sorry for the misunderstanding.Many Thanks.
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    Re: combinations

    I don't know anything about the formula you talked about. .They don't show that in school but ask in the exam.
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    Re: combinations

    I have one last question.Why did we divide by 4!?
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    Re: combinations

    There are 4!
    permutations; hence, all the ordered permutations can be grouped into
    sets of 4! assignments, and each set of 4! corresponds to a single
    unordered partition.
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