* * * can form 9*9*8 = 648. (Note that the percentile cannot be 0)
Then divide 1-9 into three groups.
A: 1 4 7
B: 2 5 8
C: 3 6 9
First, calculate how many numbers that are composed of these nine numbers and can be divisible by 3.
Any permutation of these three numbers in group A can form a number divisible by 3.
* * * There is p (3,3) = 3 * 2 *1= 6.
Similarly, Group B and Group C can each form 6 numbers divisible by 3.
All the numbers above * * * 18 can be divisible by 3.
Randomly select 1 number from groups A, B and C, and all three numbers can form a number divisible by 3.
* * * There are c (31) * c (31) * c (31) * p (3,3) = 3 * 3 * 2 *1= 65438+.
The above total * * * has18+162 =180 numbers divisible by 3.
Let's consider three digits containing 0.
Two numbers are randomly selected from Group C to form a three-digit number divisible by 0 and 3.
* * * has c (3 3,2 2) * [p (3 3,3)-p (2,2)] = 3 * (3 * 2 *1-2 *1) =12.
They are 306 309 360 390 603 609 630 690 903 906 930 960 respectively.
C (3,2) represents the number of methods for extracting two numbers from three numbers of 369 * * *.
P (3,3) indicates the number of categories in which 0 is arranged by the two numbers * * * extracted above.
P (2,2) indicates the number of categories when 0 ranks among the hundreds in the above arrangement.
Any one of Group A and Group B can use 0 to form a three-digit number divisible by 3.
* * * has c (3, 1) * c (3,1) * [p (3,3)-p (2,2)] = 3 * 3 * (3 * 2 *1-2 */kloc-.
The above total is 180+ 12+36=228. In other words, 228 of the 648 numbers are divisible by 3. The number of numbers that are not divisible by 3 is 648-228 = 420.
So the probability is: 420/648 = 35/54.