<s>[INST] 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?
 [/INST] 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.
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