Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.

What is the probability that he wears all three shirts after 6 days?

My try: 
$$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$
First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.
Thanks in advance.

Time to use inclusion-exclusion: the probability that he'll use all three is the probability that the sequence cannot come from a corpus of just a given two.
So we have: $$P(\text {all of }RGW)=P(RGW)-P(RG)-P(RW)-P(GW)+P(R)+P(G)+P(W)-P(\emptyset)$$.  Which is relatively simple: each $P$ is the summed probability of getting a shirt that's in the list, to the sixth power (because six days).
Replacing and doing the obvious simplification of using a common denominator: $$\begin {align}&\frac{6^6-5^6-4^6-3^6+3^6+2^6+1^6-0^6}{6^6}\\=&\frac{46656-15625-4096-729+729+64+1-0}{46656}\\=&\frac{27000}{46656}\\=&\frac{125}{216}
\end {align}$$
... Which I'll be honest is a much nicer number then I expected
