Ivan has 3 red blocks, 4 blue blocks and 2 green blocks. He builds the tower with randomly selected blocks but only stops when the tower consists of all three colours.

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*What is the probability that the tower is 4 blocks tall?

My approach is the following but I am not sure at all:
to make it 3 colours it builds a tower that is either 3 blocks or 8 blocks.
8 blocks would happen in the case where the first 7 blocks are 4Blue and 3Red, so the 8th must be the green block.
so $E(8) = 4B,3R,1G = \binom{4}{4}*\binom{3}{3}*\binom{2}{1}= 2$
$E(7) = 4B,2G,1R = \binom{4}{4}*\binom{2}{2}*\binom{3}{1}= 3$
E(6) = let's reason the other way around. First $\binom{9}{6} = 84$. then the combinations of 6 blocks with only 2 colours is 4B,2R or 4B,2G  or  3B,3R.
This makes:
$ \binom{4}{4}*\binom{3}{2} + \binom{4}{4}*\binom{2}{2}+ \binom{4}{3}*\binom{3}{3}= 8$.
So $E(6) = 84 - 8 = 76$
E(5) = 3B,1R,1G  or  3R,1B,1G  or  2B,2R,1G  or  2B,2G,1R  or  2R,2G,1B =
$E(5) = \binom{4}{3}*\binom{3}{1}*\binom{2}{1}+ \binom{3}{3}*\binom{4}{1}*\binom{2}{1}+ \binom{4}{2}*\binom{3}{2}*\binom{2}{1}+ \binom{4}{2}*\binom{2}{2}*\binom{3}{1}+ \binom{3}{2}*\binom{2}{2}*\binom{4}{1}= 98$
E(4) = 2B,1R,1G  or  2R,1B,1G  or  2G,1B,1R =
$\binom{4}{2}*\binom{3}{1}*\binom{2}{1}+ \binom{3}{2}*\binom{4}{1}*\binom{2}{1}+ \binom{2}{2}*\binom{4}{1}*\binom{3}{1}= 72$
$E(3) = \binom{4}{1}*\binom{3}{1}*\binom{2}{1}= 24$
total combinations $(T) = 24 + 72 + 98 + 76 + 3 + 2 = 275$.
The probability of stopping at 4 blocks is $E(4)/(T) = 72/275$
but this is a very long solution. Do you think that is correct? if so, is there any other method to make it shorter?
 A: There are three cases to consider.
Case 1: The fourth block is green.
Of the first three blocks, none are green $\binom{7}{3}$, and they are not all blue $\binom{4}{3}$ or all red $\binom{3}{3}$. Then the probability of the fourth block being green is $\frac{2}{6}$
$$\frac{\binom{7}{3}-\binom{4}{3}-\binom{3}{3}}{\binom{9}{3}}\times \frac{2}{6}=\frac{10}{84}$$
Case 2: The fourth block is red.
Of the first three blocks, none are red $\binom{6}{3}$, and they are not all blue $\binom{4}{3}$. Then the probability of the fourth block being red is $\frac{3}{6}$
$$\frac{\binom{6}{3}-\binom{4}{3}}{\binom{9}{3}}\times \frac{3}{6}=\frac{8}{84}$$
Case 3: The fourth block is blue.
Of the first three blocks, none are blue $\binom{5}{3}$, and they are not all red $\binom{3}{3}$. Then the probability of the fourth block being blue is $\frac{4}{6}$
$$\frac{\binom{5}{3}-\binom{3}{3}}{\binom{9}{3}}\times \frac{4}{6}=\frac{6}{84}$$
The total probability is $$\frac{10+8+6}{84}=\frac{24}{84}=\frac{2}{7}$$
A: We can easily solve using generating functions
For blocks of $4$ that have all three colors, the number of permutations is given by
coefficient of $x^4$ in $4!(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!})(x+\frac{x^2}{2!} + \frac{x^3}{3!})(x+\frac{x^2}{2!}) = 36$
while unrestricted permutations = $\dfrac{9!}{4!3!2!} = 1260$
Thus $Pr = \dfrac{36}{1260} = \dfrac{2}{7}$
