How many distinct combinations of five marbles could be drawn? A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations of five marbles could be drawn?
(A) 41
(B) 51
(C) 62
(D) 72
(E) 82
I am lost in this question, can someone help?
 A: Consider cases:


*

*From the seven different colors of marbles, choose five distinct ones.

*Select two red marbles or two green marbles, then choose three distinct marbles from the other six colors.

*Choose two red marbles and two green marbles, then choose one marble from the other five colors.

*Choose three green marbles and two distinct marbles from the other six colors.

*Choose three green marbles and two red marbles.


Keep in mind that the green marbles are identical so we do not care which green marbles are selected.  The same observation holds for the red marbles.

 $$\binom{7}{5} + \binom{2}{1}\binom{6}{3} + \binom{5}{1} + \binom{6}{2} + 1$$

A: Let $k$ be the number of green and red marbles chosen, and let $n$ be the number of other marbles chosen.  
Then $k+n=5$, and if $g(k)$ is the number of ways to choose the $k$ marbles,
then $g(0)=g(5)=1,\; \;g(1)=g(4)=2,\;\;g(2)=g(3)=3$.
Since there are $\dbinom{5}{n}$ ways to choose the other $n$ marbles,
there are $\displaystyle1\binom{5}{5}+2\binom{5}{4}+3\binom{5}{3}+3\binom{5}{2}+2\binom{5}{1}+1\binom{5}{0}=82$ selections.

Alternatively, we can find the number of solutions of $x_1+\cdots+x_7=5$ in nonnegative integers
$\hspace .4 in$where $\;x_1\le3, \;x_2\le2,\;$ and $\;x_i\le1\;$ for $3\le i\le 7$.
If we use Inclusion-Exclusion, 
letting $E_1$ be the selections with $x_1\ge4$, $E_2$ the selections with $x_2\ge3$, and 
$\hspace .4 in E_i$ the selections with $x_i\ge2$ for $3\le i\le7$,
this gives $\binom{11}{6}-\binom{7}{6}-\binom{8}{6}-5\binom{9}{6}+5\binom{6}{6}+\binom{5}{2}\binom{7}{6}=82$ choices.
