Use a total of 6 numbers from 0 to 5 to form 6 digits larger than 240135. How many are there? I hope to give a detailed answer (using permutation and combination)

Use a total of 6 numbers from 0 to 5 to form 6 digits larger than 240135. How many are there? I hope to give a detailed answer (using permutation and combination)

The first step, if the number is larger than 240135, then the first digit must be greater than 2. There are only three possibilities: 3, 4 and 5. Other numbers do not need to be arranged in order, that is, 3 * 5! The second step, if the first digit is 2, then the second digit may be 4 or 5. If it is 5, the remaining numbers are 0, 1, 3 and 4

How many kinds of five different numbers can be randomly arranged and combined into five digits?

If there is no 0, the result is a full permutation of 5:
5!=120；
If there is 0, then 0 cannot be put in the first place
5!- 4!=120-24
=96

Write the number 123456 on the six sides of a small cube
On the six sides of a small cube, write 1, 2, 3, 3 respectively in any order. Throw it 50 times. Count the number of times 1, 2, 3 face up. Then calculate the percentage of the number of times each number faces up in the total number of times
What is the law?

This is a problem of probability. The numbers 1, 2 and 3 are about 1 * 50 / 6, 2 * 50 / 6 and 3 * 50 / 6, respectively. The percentage of the number of each number up in the total throwing times is 1 / 6 * 100%, 2 / 6 * 100% and 3 / 6 * 100%
The rule is: the number of possible upward / the total number of possible upward