In how many ways could six of the colored Easter eggs been crushed? Problem:
Mother was carrying a basket with 11 red, 8 blue, 5 green and 5 yellow decorated Easter eggs (eggs
of the same color are identical). 
When entering the room of her son Thomas, she tripped
over door step and crushed six eggs. 
Thomas paused in thought for a moment, and than he said to
his mom the number expressing, all the options of which various colors six eggs could be crushed.
Determine the number given by Thomas.
Here what have I done so far:
$$Omega = ^{29}C_6$$


*

*The first case (2 red, 2 blue, 1 green, 1 yellow)


We have : $$^{11}C_2\cdot ^8C_2\cdot ^5C_1\cdot ^5C_1$$ 


*

*The second case (2 red, 1 blue, 2 green, 1 yellow)


We have : $$^{11}C_2\cdot ^8C_1\cdot ^5C_2\cdot ^5C_1$$ 


*

*The third case (2 red, 1 blue, 1 green, 2 yellow)


We have : $$^{11}C_2\cdot ^8C_1\cdot ^5C_1\cdot ^5C_2$$ 


*

*The fourth case (1 red, 2 blue, 2 green, 1 yellow)


We have : $$^{11}C_1\cdot ^8C_2\cdot ^5C_2\cdot ^5C_1$$ 


*

*The fifth case (1 red, 2 blue, 1 green, 2 yellow)


We have : $$^{11}C_1\cdot ^8C_2\cdot ^5C_1\cdot ^5C_2$$ 


*

*The sixth case (1 red, 1 blue, 2 green, 2 yellow)


We have : $$^{11}C_1\cdot ^8C_1\cdot ^5C_2\cdot ^5C_2$$ 


*

*The seventh case (3 red, 1 blue, 1 green, 1 yellow)


We have : $$^{11}C_3\cdot ^8C_1\cdot ^5C_1\cdot ^5C_1$$ 


*

*The eighth case (1 red, 3 blue, 1 green, 1 yellow)


We have : $$^{11}C_1\cdot ^8C_3\cdot ^5C_1\cdot ^5C_1$$ 


*

*The ninth case (1 red, 1 blue, 3 green, 1 yellow)


We have : $$^{11}C_1\cdot ^8C_1\cdot ^5C_3\cdot ^5C_1$$ 


*

*The tenth case (1 red, 1 blue, 1 green, 3 yellow)


We have : $$^{11}C_1\cdot ^8C_1\cdot ^5C_1\cdot ^5C_3$$ 
I have read the problem so many time, and have some confusion about "various colors six eggs", and I feel something wrong with my solution.
I do not know if all those case I have done above are enough, if not please show step if you know how to do it.
 A: Trying to list all the cases is inadvisable.  Also, since eggs of the same color are considered to be indistinguishable, what matters here is how many eggs of each color were crushed, not which eggs were crushed.
Let $x_r$, $x_b$, $x_g$, and $x_y$ denote, respectively, the number of red, green, blue, and yellow eggs that were crushed.  Since a total of six eggs were crushed,
$$x_r + x_b + x_g + x_y = 6 \tag{1}$$
which is an equation in the nonnegative integers subject to the restrictions that $x_r \leq 11$, $x_b \leq 8$, $x_g \leq 5$, and $x_y \leq 5$.  Observe that it is only possible to violate the last two restrictions.
A particular solution of equation 1 corresponds to the placement of three addition signs in a row of six ones.  For instance,
$$1 1 + 1 + + 1 1 1$$
means $x_r = 2$, $x_b = 1$, $x_g = 0$, and $x_r = 3$.  The number of such solutions is the number of ways we can place three addition signs in a row with six ones, which is
$$\binom{6 + 4 - 1}{4 - 1} = \binom{9}{3}$$
since we must choose which three of the nine positions required for six ones and three addition signs will be filled with addition signs.
However, these solutions include those in which the restrictions $x_g \leq 5$ or $x_y \leq 5$ are violated.  We must subtract these from the total.
If $x_g > 5$, then $x_g = 6$ and $x_r = x_b = x_y = 0$, so only one solution of equation 1 violates the restriction that $x_g \leq 5$.
By symmetry, there is also only one solution of equation 1 violates the restriction that $x_y \leq 5$, namely $x_r = x_b = x_g = 0$ and $x_y = 6$.
Hence, the number of ways six of the colored eggs that were initially in mother's basket could have been crushed is
$$\binom{9}{3} - 2 = 84 - 2 = 82$$
A: It doesn't matter which specific red, blue, green, or yellow were crushed.  Eggs the same color are considered equivalent.
Let $R = $ then number of red eggs.  $R$ can be $0$ or it can be as high as $6$.
Let $B = $ be the number of blue eggs.  $B$ can be as low as $0$ or as high as $6-R$
Let $G = $ the number of green eggs.  $G$ can be as low as $0$ or as high as $\min(6-R-B, 5)$.  However if $R=B =0$ then as there are only $5$ yellow eggs, the would have to be at least $1$ green egg.  So $G$ can be as low as $\max(0, 1-R-B)$..
Let $Y=$ the number of yellow eggs.  $Y = 6-R-B-G$.  Once $R,B,G$ are determined there is only $1$ value of this.
Number ways this can be is:
$\sum\limits_{R=0}^6 \sum\limits_{B=0}^{6-R}\sum\limits_{G=\max(0, 1-R-B)}^{\min(6-R-B, 5)}1$
THose conditionals make it hard so I think we should divide it into cases
Case 1:  $R = 0; B=0$.
Then the number of ways to do it is:
$\sum\limits_{R=0}^0 \sum\limits_{B=0}^{0}\sum\limits_{1}^{5}1=$
$ \sum\limits_{1}^{5}1=5$.
Case 2: There is at least one red or blue eggs.
Case 2a:  There are no red eggs but at least one blue egg.
Then the numbers of ways to do this is
$\sum\limits_{R=0}^0 \sum\limits_{B=1}^{6}\sum\limits_{0}^{5}1=$
$\sum\limits_{B=1}^{6}\sum\limits_{0}^{5}1 = $
$\sum\limits_{B=1}^{6}6 = 36$
Case 2b: There is at least one red egg 
Then the number of ways to do it are:
$\sum\limits_{R=1}^6 \sum\limits_{B=0}^{6-R}\sum\limits_{G=0}^{6-R-B}1$
Now $\sum\limits_{G=0}^{6-R-B}1= 6-R-B + 1=7-R-B$.
So the number of ways to do the eggs is:
$\sum\limits_{R=0}^6 \sum\limits_{B=0}^{6-R}(7-R-B)$
Now $\sum\limits_{B=0}^{6-R}(7-R-B)=$
$\sum\limits_{B=0}^{6-R}(7-R) - \sum\limits_{B=0}^{6-R}B =$
$(7-R)(6-R+1)- \frac {(6-R)(6-R+1)}2=$
$(7-R)^2 - \frac {(7-R)(6-R)}2 =\frac{2(7-R)^2 - (7-R)(6-R)}2=$
$\frac {(7-R)(2(7-R) - (6-R))}2=\frac {(7-R)(8-R)}2=$
$\frac {R^2 - 15R + 56}2$
So the number of ways to do the eggs if there is at least one egg is:
$\sum\limits_{R=1}^6 \frac {R^2 - 15R + 56}2=$
$\frac 12\sum\limits_{R=1}^6 R^2 -\frac {15}2\sum\limits_{R=1}^6 R + 28\sum\limits_{R=1}^61=$
$\frac 12\frac {6*7*13}6 - \frac {15}2\frac {6*7}2 + 28*6 =$
$\frac {13}2 - \frac{15*21}2 + 168 =$
$6\frac 12 - 157\frac 12 + 168 = 17$.
So there are $17+36+5 =58$ ways to do this.
....
A: The answer to this question is the coefficient of $x^6$ in the following expansion:
$$(1 - x)^{-4}(1 - x^{12})(1 - x^9)(1 - x^{6})(1 - x^6)$$
I derived this result in this answer: Integer solution for $\sum_{i=1}^rx_i=k$ where $x_i$ has bounds.
Solving further:

$(1 - x)^{-4}(1 - x^9 - 2x^{6}$ and other higher useless terms$)$

Removing all useless terms:

$(1 - x)^{-4}(1 - 2x^{6})$
$(1 - x)^{-4} - 2(1 - x)^{-4}.x^6$
Answer = $^{6 + 4 - 1}C_{4 - 1} - 2^{0 + 4 - 1}C_{4 - 1}$
= $^9C_3 - 2.{^3C_3}$
= 82

