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.
 A: The alternative problem is this: one has 3 shirts of red color, 2 of green and 1 of white of which he chooses arbitrarily. What is the probability that he puts on all of the 3 colors within 3 days?$$P=\dfrac{N}{N_{\text{tot}}}\\N=\binom{3}{1}\binom{2}{1}\binom{1}{1}=6\\N_{\text{tot}}=\dfrac{6!}{1!2!3!}=60\\P=0.1$$
A: 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
A: Let $R$ denote the event that he wears a red shirt in the $6$ days.
Let $G$ denote the event that he wears a green shirt in the $6$ days.
Let $W$ denote the event that he wears a white shirt in the $6$ days.
To be found is: $$P(R\cap G\cap W)=1-P(R^{\complement}\cup G^{\complement}\cup W^{\complement})=$$$$1-P(R^{\complement})-P(G^{\complement})-P(W^{\complement})+P(R^{\complement}\cap G^{\complement})+P(R^{\complement}\cap W^{\complement})+P(G^{\complement}\cap W^{\complement})=$$$$1-\left(\frac36\right)^6-\left(\frac46\right)^6-\left(\frac56\right)^6+\left(\frac16\right)^6+\left(\frac26\right)^6+\left(\frac36\right)^6$$
A: I will try to replicate @Mustafa's answer in a different way.
Guess there is $6$ white tshirts, $3$ green, and $2$ red. For this person to have picked one atleast from each of all of them during 6 days it sums to $\sum\binom{1}{6}^i\binom{1}{3}^j\binom{1}{2}^k$ for $i$ $j$ $k$ strictly positive and $i+j+k<=6$.
Taking count of ordering implicitely between distinct colors gives:  $6!(\frac{2*3*6^4}{4!}+\frac{2*3^2*6^3}{2!3!}+\frac{2^2*3*6^3}{2!3!}+\frac{2*3^3*6^2}{2!3!}...=84180$
All possible combinations restrictlessly of any choice sums up the previous value to $6!(\frac{6^6+3^6+2^6}{6!}+\frac{3^5(2+6)+2^5(3+6)+6^5(2+3)}{5!}+\frac{3^4(2^2+6^2)+2^4(3^2+6^2)+6^4(2^2+3^2)}{2!4!}+\frac{3^3(2^3+6^3)+6^32^3}{3!3!})$ which equals $761761$
Probability is $\frac{84180}{761761+84180} \approx 0.1$ I'm surprised it's the same amount !!
