A six digit number is formed randomly using digits $\{1,2,3\}$ with repetitions. Choose the correct option(s): 
A six digit number is formed randomly using digits $\{1,2,3\}$ with repetitions. Choose the correct option(s):

*

*A) Probability that all digits are used at least once is $\dfrac{20}{27}$

*B) Probability that digit $1$ is used odd number of times and $2$ is used even number of times is $\dfrac{1-3^{-6}}4$

*C) Probability that all digits are used as well as odd digits are used odd number of times and even digit is used even number of times is $\dfrac{3^6-2^7+1}{4\cdot3^6}$

Total cases = $3^6$

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*A) All digits used at least once = Total cases$-$Cases when one digit is not used

Or, would it be
All digits used at least once = Total cases$-$Cases when one digit is not used$-$Cases when two digits are not used
In any case, I am not getting $\frac{20}{27}$

*

*B) $1$ can be used $1$ or $3$ or $5$ times. $2$ can be used $0$ or $2$ or $4$ times.

So, favorable cases=$^6C_1(^5C_0+^5C_2+^5C_4)+^6C_3(^3C_0+^3C_2)+^6C_5=182$, and it matches with the answer.
But the answer given in the option is of different format $\dfrac{3^6-1}{4\cdot3^6}$. Looks like they are subtacting $1$ from the total cases and diving by $4$ to get the favorable case. Why?

*

*C) $1$ can be used $1$ or $3$ or $5$ times. $2$ can be used $2$ or $4$ times.

So, favorable cases=$^6C_1(^5C_2+^5C_4)+^6C_3(^3C_2)+^6C_5=156$
But as per the option, favorable cases= $150.5$. Also, looks like they are subtracting $2^7-1$ from total cases and then diving by $4$, what could be the motivation for this, even if this is wrong?
 A: For the first, it is application of Principle of Inclusion Exclusion or we can simply work as follows -
Total count of $6$ digit numbers $ = 3^6$
Count of numbers where one of the digits is missing $ = 3 \cdot (2^6 - 2)$
Count of numbers where two digits are missing $ = 3$
So the answer for $(a)$ is $ = \dfrac{3^6 - 3 \cdot 2^6 + 3}{3^6} = \dfrac{20}{27}$
For the second, your working is correct but another way to solve it would be,
Count of $6$ digit numbers where $1$ appears odd number of times and $2$ appears even number of times,
$ \displaystyle {6 \choose 1}\frac{2^5}{2} + {6 \choose 3}\frac{2^3}{2} + {6 \choose 5} = 182$
Explanation: Take the first term where $1$ occurs once. We choose a place for it from $6$ digits. Rest $5$ digits are made up of $2$ and $3$. In half of the numbers, $2$ will occur even number of times $(0, 2, 4$ times) and in other half it will occur odd number of times $(1, 3, 5$ times). Similarly other terms.
Now given first and second are correct. It must be third that is incorrect. Now notice that the third is a subset of second. In second, as $1$ occurs odd number of times, $2$ occurs even number of times and so $3$ occurs odd number of times. But in third, we need to subtract cases from second where $2$ was missing (occurred zero times). Can you take it from here?
A: +1 to your answer, for showing your work.  I agree that your analysis for Part B is correct.  My analysis is:
Part A
Use Inclusion-Exclusion.
$\displaystyle \frac{N\text{(umerator)}}{D\text{(enominator)}},~$ with $\displaystyle D = \frac{1}{3^6}.$
$\displaystyle N = 3^6 - \left[\binom{3}{1}2^6\right] + \left[\binom{3}{2}1^6\right] = 729 - 192 + 3 = 540.$
$\displaystyle \frac{540}{729} = \frac{20}{27}$ 
So, the answer given for part (A) is correct.

Part B
Again, $\displaystyle D = 3^6.$
$\displaystyle N = \sum_{k=1}^6 f(k),~$ as described below:
$f(1) : (1)$ "1", $(0)$ 2's : $\binom{6}{1} = 6.$ 
$f(2) : (1)$ "1", $(2)$ 2's : $\binom{6}{1}\binom{5}{2} = 60.$ 
$f(3) : (1)$ "1", $(4)$ 2's : $\binom{6}{1}\binom{5}{4} = 30.$ 
$f(4) : (3)$ "1"'s, $(0)$ 2's : $\binom{6}{3}\ = 20.$ 
$f(5) : (3)$ "1"'s, $(2)$ 2's : $\binom{6}{3}\binom{3}{2} = 60.$ 
$f(6) : (5)$ "1"'s, $(0)$ 2's : $\binom{6}{5} = 6.$ 
$\displaystyle N = 182.$
$\displaystyle \frac{1 - (1/729)}{4} = \frac{728}{4 \times 729} = \frac{182}{729}.$
So, the answer given for part (B) is also correct.

Part C 
Again, $\displaystyle D = 3^6.$
$\displaystyle N = \sum_{k=1}^3 f(k),~$ as described below:
$f(1) : (1)$ "1", $(1)$ "3", $(4)$ 2's : $\binom{6}{1}\binom{5}{1} = 30.$ 
$f(2) : (1)$ "1", $(3)$ "3"'s, $(2)$ 2's : $\binom{6}{3}\binom{3}{2} = 60.$ 
$f(3) : (3)$ "1's", $(1)$ "3", $(2)$ 2's : $\binom{6}{3}\binom{3}{2} = 60.$ 
$\displaystyle N = 150.$
$\displaystyle \frac{3^6 - 2^7 + 1}{4 \times 3^6} = \frac{602}{4 \times 729} \neq \frac{150}{729}.$
So, the answer given for part (C) is wrong.
Note that if the answer given for part (C) had been 
$\displaystyle \frac{3^6 - 2^7 - 1}{4 \times 3^6}$, then the answer would have been correct.
A: From @MathLover and @user2661923 answers, I got an idea to try to approach the solution in a way that would match the format of the options. So, here is my solution:
For part B), if $1$ is to come odd number of times, it would be $50\%$ of the total cases. And $2$ would come even number of times in $50\%$ of those cases. So, overall $25\%$ of the total cases. But here, we have also included a case where $2$ has come six times, and probability of that case is $\dfrac1{3^6}$. This needs to be subtracted i.e. $1-3^{-6}$. Now, taking its $25\%$, we get $\dfrac{1-3^{-6}}4$, which is the format of the option.
From this, we need to subtract those cases where $2$ had come zero times i.e. in those cases, only $1$ and $3$ could be filled. But we would consider only those cases where $1$ had come odd number of times, because those are the cases we are dealing with here. So, cases = $\dfrac{2^6}2$. Probability=$\dfrac{2^6}{2\cdot3^6}$.
So, the required probability for C) part$$=\dfrac{1-3^{-6}}4-\dfrac{2^6}{2\cdot3^6}\\=\frac{3^6-1}{4\cdot3^6}-\frac{2^7}{4\cdot3^6}\\=\frac{3^6-2^7-1}{4\cdot3^6}$$
