There is a lottery game , it consists of $6$ block and each blocks contains numbers from $1$ to $90$.When you play the lottery , you should select one number from each of these $6$ blocks.
$\color{red}{a-)}$ What is the probability of winning lottery if winner numbers are different from each other (i.e repetition is not allowed) and the order of numbers matters?
(for example if winner lottery is $5-12-45-85-3-2$ respectively, you cannot take money if your numbers are $12-5-45-85-3-2$ ,respectively.)
$\color{red}{b-)}$ What is the probability of winning lottery if winner numbers can repeat and the order of numbers are matter ?
$\color{red}{c-)}$ What is the probability of winning lottery if winner numbers are different from each other (i.e repetition is not allowed) and the order of numbers does not matter ?
(For example , if the seleced numbers are $3-4-56-78-46-33$ , then the tickets have any permutation of these numbers will be winner such that $3-4-56-78-46-33$ ,$56-78-46-3-4-33$ ,$4-78-33-46-56-3$ ,etc are winners)
$\color{red}{d-)}$ What is the probability of winning lottery if winner numbers can repeat and the order of numbers does not matter ?
(For example , if the winner numbers are $2-19-2-89-89-23$, then any permutation of them can take prize such as $2-19-2-89-89-23$ ,or $19-2-2-23-89-89$ ,or $89-2-23-89-2-19$ )
MY ATTEMPT:
$\color{blue}{a-)}$ $\frac{1}{P(90,6)}$
$\color{blue}{b-)}$ $\frac{1}{90^{6}}$
$\color{blue}{c-)}$ $6! \times \frac{1}{P(90,6)}$
$\color{blue}{d-)}$ I thought that i should make use of exponential function such as $(e^x)^6 $ and find the coefficient of $\frac{x^6}{6!}$ , so the answer is $\frac{6^6}{90^6} $ or should i use $(e^x)^{90} $ and find the coefficient of $\frac{x^6}{6!}$ , so the answer is $\frac{90^6}{90^6}=1 $
Is my solutions correct ? If not , can you help to correct them ?
EDITED NOTE = selection probability of each number is equal