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Math Help - Number of ways in which 7 distinct objects can be put..?

  1. #1
    Super Member fardeen_gen's Avatar
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    Number of ways in which 7 distinct objects can be put..?

    Find the number of ways in which 7 distinct objects can be put in three identical boxes so that no box remains empty.
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  2. #2
    Super Member Deadstar's Avatar
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    This is similar to a problem i had in an old course where you had to find integers x,y,z s.t. x + y + z = some +ve integer.

    So i think this one can be reworded as the 3 boxes being x,y and z. And n being 7 distinct objects.

    Since you want +ve integers what to do is set x' = x-1 etc and then you get x' + y' + z' = 4.

    This is done by using the binomial expression \left(\begin{array}{cc}n + k-1\\k-1\end{array}\right) where k is the number of 'boxes'

    So we get \left(\begin{array}{cc}4+3-1\\3-1\end{array}\right) = \left(\begin{array}{cc}6\\2\end{array}\right) = 15.
    Last edited by Deadstar; May 27th 2009 at 06:29 PM.
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Deadstar View Post
    This is similar to a problem i had in an old course where you had to find integers w,x,y,z s.t. w + x + y + z = some +ve integer.

    So i think this one can be reworded as the 4 boxes being w,x,y and z. And n being 7 distinct objects.

    Since you want +ve integers what to do is set w' = w-1 etc and then you get w + x + y + z = 5.

    This is done by using the binomial expression \left(\begin{array}{cc}n + k-1\\k-1\end{array}\right) where k is the number of 'boxes'

    So we get \left(\begin{array}{cc}5+4-1\\4-1\end{array}\right) = \left(\begin{array}{cc}8\\3\end{array}\right) = 56.
    its  3 boxes.
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  4. #4
    Super Member Deadstar's Avatar
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    Ah grim. Fixed!
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  5. #5
    Super Member Deadstar's Avatar
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    Actually my first post was totally full of fail thanks for noticing that boxes thing...
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  6. #6
    Super Member fardeen_gen's Avatar
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    The answer is 322. Are we missing something?
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  7. #7
    Super Member Deadstar's Avatar
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    Yeah i think my way was if there was 7 identical objects... 15 is waaaay to low.
    But how did you get 322? I just did a 'brute force' method and came up about 99 but didnt really check it.
    Since each box must contain 1 object it can be simplified down to how many ways can 4 objects be distributed between 3 boxes... This questions bugging me now...

    EDIT: Is the answer just 3^4 = 81?
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  8. #8
    Senior Member pankaj's Avatar
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    Well,this is how it goes.

    Since the boxes are identical the problem reduces to the fact 7 distinct objects have to be divided(not distributed) into 3 groups as arrangement among the boxes are immaterial since the boxes are identical

    The groupings can be done as follows 1,3,3),(1,2,4),(2,2,3),(1,1,5)" alt="1,3,3),(1,2,4),(2,2,3),(1,1,5)" />

    This can be done in following ways:
    = \frac{7!}{1!3!3!}\frac{1}{2!}+\frac{7!}{1!2!4!}+\f  rac{7!}{2!2!3!}\frac{1}{2!}+\frac{7!}{1!1!5!}\frac  {1}{2!}

    =70+105+105+21

     <br />
=301<br />

    Either the answer is 301 or I have missed a grouping which is not likely
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  9. #9
    Senior Member Sampras's Avatar
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    Quote Originally Posted by pankaj View Post
    Well,this is how it goes.

    Since the boxes are identical the problem reduces to the fact 7 distinct objects have to be divided(not distributed) into 3 groups as arrangement among the boxes are immaterial since the boxes are identical

    The groupings can be done as follows 1,3,3),(1,2,4),(2,2,3),(1,1,5)" alt="1,3,3),(1,2,4),(2,2,3),(1,1,5)" />

    This can be done in following ways:
    = \frac{7!}{1!3!3!}\frac{1}{2!}+\frac{7!}{1!2!4!}+\f  rac{7!}{2!2!3!}\frac{1}{2!}+\frac{7!}{1!1!5!}\frac  {1}{2!}

    =70+105+105+21

     <br />
=301<br />

    Either the answer is 301 or I have missed a grouping which is not likely
    What about  (1,4,2) ? Or  (1,5,1) ? Also why did you divide by  2! ?
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  10. #10
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    301 is correct.

    The number is also S(7,3), where S is the Stirling Number of the 2nd Kind.
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  11. #11
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Sampras View Post
    What about  (1,4,2) ? Or  (1,5,1) ? Also why did you divide by  2! ?
    (1,2,4) is same as (1,4,2) since boxes are identical.There is no such thing as the first box or the second or the third .

    If 2 groups contain same number of objects then it is required that we divide by 2!.
    For example if {a,b,c,d,e,f,g} are the 7 obects.One grouping containing (2,2,3) objects can be described as follows.
    {a,b},{c,d}{e,f,g} and another can be {c,d},{a,b}{e,f,g}.
    Now arent these groupings same.
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