Question 1204839
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There is a group of five children, where two of the children are twins. In how
many ways can I distribute 
18 identical pieces of candy to the children, if the twins must get 7 pieces of candy?
~~~~~~~~~~~~~~~~~~~~~



I came with third interpretation, which differs from that by Edwin's and @greenestamps.



I will assume that


    (1)    for twins and for other 5-2 = 3 children the distributions allow 0 pieces for some child/children
     
                 (but the sum of pieces for twins is always 7 and the sum of pieces for other three children is always 18-7 = 11)


    (2)    The number of pieces each twin gets is IMPORTANT.



So,  my interpretation is half-way between that of  Edwin and @greenestaps.


In support of this interpretation,  I should say following:


<pre>
    (a)  My assumption (1) is inhumane; but this problem is not about humanitarian mission - it is about
         solving Math problem as it is given.  So, I leave humanitarian reasons aside as irrelevant.

    (b)  My assumption (2) reflects the fact that the twins are two different distinguishable individuals;
         therefore, when they obtain different numbers of pieces, it corresponds to different distinguishable distributions.
</pre>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;After all these explanations, below is the solution itself.



<pre>
So, for 3 children, I have the "stars and bars" distribution problem with n=3 persons and k=11 pieces, with 0 (zero)
lower boundary of pieces for individuals.

For this 3 children, the number of possible distributions is {{{C[n+k-1]^(n-1)}}} = {{{C[3+11-1]^(3-1)}}} = {{{C[13]^2}}} = {{{(13*12)/2}}} = 78.


For the twins, there are 7+1 = 8 different distributions (they correspond to  numbers {0,1,2,3,4,5,6,7} of pieces 
which one of the twins does obtain; then the other twin gets the rest of 7).


So, the <U>ANSWER</U>  to the problem is this product  8*78 = 624.
</pre>

This is my solution in my interpretation.