当两个骰子同时掷出时，结果之和为素数的概率是多少？
在数学中，排列是指组织集合的行为，其中集合的所有成员都排列成某种顺序或顺序。换句话说，如果集合已经被排列，那么它的组件的重新排列称为置换过程。几乎所有数学领域都以或多或少的重要方式发生排列。当考虑某些有限集上的不同命令时，它们经常出现。
什么是组合？
组合是从组中选择项目的行为，这样（与排列不同）选择顺序无关紧要。在较小的情况下，可以计算合并的数量。组合是指一次取 k 的 n 样东西不重复地合并。为了指代允许重复出现的组合，经常使用术语 k-selection 或 k-combination with repeat。
置换公式
排列是从一组 n 个事物中选择 r 个事物而无需替换，并且顺序很重要。
n P r = (n!)/(nr)!
组合配方
组合是从 n 个事物的集合中选择 r 个事物而无需替换且顺序无关紧要。
当两个骰子同时掷出时，结果之和为素数的概率是多少？
解决方案：
The total number of possible outcomes in this case is 36-
{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),
(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),
(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
Out of the listed possibilities, the favorable outcomes are-
{(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),
(6,5)}
Hence, the number of favorable outcomes is 15.
Therefore, the required probability is 15/36 = 5/12
类似问题
问题1：掷三个骰子时，如果没有两个可能相同，可能有多少结果？
解决方案：
The first two dice together have 6×5=30 possible outcomes, from above.
For each of these 30 outcomes, the third die has four possible outcomes,
So the total number of outcomes is 30×4=6×5×4=120. (Note that we consider the dice to be detectable,
that is, a roll of 6, 4, 1 is unlike than 4, 6, 1, because according to the question the first and second dice are different in the two rolls, even though the numbers as a place are similar.)
问题 2：假设编号为 1 到 n 的块在桶中；我们拉出其中的 k 个，像我们一样将它们排成一行。有多少结果可能？也就是说，我们可以看到多少种不同的 k 块排列？
解决方案：
This is essentially the same as the previous question: there are k “spots” to be filled by blocks. Any of the n blocks might seem first in the line; then any of the remaining n−1 might seem next, and so on. The number of outcomes is thus
n(n−1)(n−2)⋯(n−k+1), by the multiplication principle. In the previous question, the first “spot” was die number one, the second spot was die number two, the third spot die number three, and 6⋅5⋅4=6(6−1)(6−2); notice that 6−2=6−3+1