Form a Committee Of 10 Senators The problem I'm working on reads: "How many different committees of 10 Senators can be formed if the two Senators from the same state, (50 States in All) are considered identical?"
This is the answer I got but I'm almost sure it's wrong:
$e_{1-50}=(1+x+x^2)$
$g(x)=(1+x+x^2)^{50}$
$(1+x+x^2)^{50}=\frac{(1-x^3)^{50}}{(1-x)^{50}}$
$=(1-x^3)^{50}*\frac{1}{(1-x)^{50}}$
$(1-x^3)^{50}=(1+(-1)(x^3 ))^{50}$
$=(1-C(50,1) x^3+C(50,2) x^6-C(50,3) x^9+⋯)$
$\frac{1}{(1-x)^{50}} =(1+C(50,1)x+C(51,2) x^2+⋯+C(59,10) x^{10}+⋯)$
$g(x)=(1-C(50,1) x^3+C(50,2) x^6-C(50,3) x^9+⋯)
   *(1+C(50,1)x+C(51,2) x^2+⋯+C(59,10) x^{10}+⋯)$
$a_{r}=C(59,10)-C(50,1)*C(56,7)+C(50,2)*C(53,4)-C(50,3)*C(50,1)=51,590,216,930$
I'm wondering where I went wrong. Any help would be greatly appreciated!
 A: We can count this by considering the number of states $i$ that contribute two senators:
$$
\sum_{i=0}^{5}\binom{50}{i}\binom{50-i}{10-2i}=51590216930.
$$
A: i think i might have a solution.  you have at least $5$ states and at most $10$ with the $2$ senators in each state considered identical.  if it is $5$ states there are none left to choose otherwise there are $2,4,6,8,10$ states selected that have $1$ senator chosen.
${5\choose 0}{50\choose 5}+{6\choose 2}{50\choose 6}+{7\choose 4}{50\choose 7}+{8\choose 6}{50\choose 8}+{9\choose 8}{50\choose 9}+{10\choose 10}{50\choose 10}$
A: I wonder at the interpretation of the question.
If there were 50 pairs of counters of different colors, I can understand,
but how can you have a committee with identical people ?
I would think the answer is simply $\dbinom{50}{10}$
Remarks
Since you were asked to use a G.F. your interpretation and that of others must be right, but I am mystified why the questioner had to insert  "the two Senators from the same state, (50 States in All) are considered identical" 
