Unordered Sample With Repetition Out of 19 different choices, I am supposed to choose 25 items. This is ${\binom{19+25-1}{25}} = {\binom{43}{25}}$. However, if two of the items cannot be chosen with repetition, how do I solve this? 
It could be ${\binom{17+23-1}{23}} + {\binom{19}{2}}$ but I dont think that is quite right. 
${\binom{17+23-1}{23}}*19*18$ could maybe be a possibility? I think this number is larger than the original though so that can't be right.
 A: Hint:  The inhomogeneity makes it messier.  I would approach it by adding the ways to do the choosing without either of the two special items, with one of the two special items, and with two of them.  With neither, you see that it is $\binom{17+25-1}{25}$ With one, you can choose the one in $2$ ways, then choose the other $24$ in how many ways (out of the $17$?).  And with two of the special ones?
Added:  You have three cases:  you choose neither of $18,19$, you choose one of them, or you choose both.  If you choose neither, there are $\binom {17+25-1}{25}$ possibilities, by the same logic as you indicate for unrestricted selection of $19$ in your question.  If you choose one, you have $2$ choices for which one, then have to select $24$ without restriction from the first $17$.  This gives $2\binom{17+24-1}{24}$ ways.  Finally, you can choose a set with both of $18,19$ in $\binom {17+23-1}{23}$ ways.  The final answer is the sum of these, as the possibilities are disjoint, for a total of  $\binom {17+25-1}{25}+2\binom{17+24-1}{24}+\binom {17+23-1}{23}$
A: There are 3 choices to choose 2 objects with no repetition: 
   a1) Choose 0 from 2, 
   b1) choose 1 from 2, 
   b1) and choose 2 from 2.
Multiply accordingly with 3 choice to choose 
   a2) 25 objects from (17-2) choices with repetition,
   b2) 24 objects from (17-2) choices with repetition,
   b2) 23 objects from (17-2) choices with repetition
Possible outcomes is,
   (a1 x a2) + (b1 x b2) + (c1 x c2)
