What is the probability of integer 1 occurring only once in a 6-digit number generated from 1, 2, 3? So seeing that the six digits number is generated from integers 1, 2, 3.
What is the probability that:


*

*Integer 1 will occur once?

*Integer 2 - 2 times?

*Integer 3 - 3 times?


I know that all possible combinations of 1, 2, 3 is given by:
$3^6 = 729$
Plus, what is the probability that an even number will be obtained from only one digit 2?
 A: Integer 1 will occur once?
$$\ 6 * 2^5 / 3^6 = 192/729 $$
There are six places for the 1 to be, and five places to be filled with combinations of the other two numbers.  So I multiply 6 with $2^5$ then divide by the number of possible combinations.
Integer 2 - 2 times?
$$\ (_6P_2 * 2^4) / (3^6 * 2!) = 240/729 $$ 
To find the number possible places the twos can be we get all permutations of $_6P_2$, then divide by $2!$ because the twos are identical, so we need to remove all concurrences of them switching places. There are now four places to be filled with combinations of the other two numbers, so we multiply by $2^4$.  Then we divide by total number of combinations.
Integer 3 - 3 times? 
$$\ (_6P_3 * 2^3) / (3^6 * 3!) = 160/729 $$ 
To find the number possible places the threes can be we get all permutations of $_6P_3$, then divide by $3!$ because the threes are identical, so we need to remove all concurrences of them switching places. There are now three places to be filled with combinations of the other two numbers, so we multiply by $2^3$.  Then we divide by total number of combinations.
