唐僧的徒弟有多少个？
user
唐僧有四个徒弟，分别是孙悟空、猪八戒、沙僧和白龙马。
assistant
关羽和张飞是师兄弟吗？
user
关羽和张飞并不是师兄弟。他们是刘备的结义兄弟，三人一起在桃园结义，共同辅佐刘备建立了蜀汉政权。在《三国演义》中，他们被称为“刘关张”。
assistant
balls into bins question. If I have 12 balls and 5 bins, what is the probability that no bin is empty?
user
To determine the probability that no bin is empty when you have 12 balls and 5 bins, we can use the principle of inclusion-exclusion.

1. **Total number of ways to distribute 12 balls into 5 bins:**
   Each ball can go into any of the 5 bins, so there are \(5^{12}\) total ways to distribute the balls.

2. **Number of ways to distribute the balls such that no bin is empty:**
   We need to count the number of ways to distribute 12 balls into 5 bins such that no bin is empty. This is a classic problem that can be solved using the principle of inclusion-exclusion.

   Let \(A_i\) be the event that the \(i\)-th bin is empty. We want to find the number of ways to distribute the balls such that none of the bins are empty, which is the complement of the union of the events \(A_1, A_2, A_3, A_4, A_5\).

   Using the principle of inclusion-exclusion, the number of ways to distribute the balls such that no bin is empty is:
   \[
   \left| \bigcup_{i=1}^5 A_i^c \right| = 5^{12} - \binom{5}{1} \cdot 4^{12} + \binom{5}{2} \cdot 3^{12} - \binom{5}{3} \cdot 2^{12} + \binom{5}{4} \cdot 1^{12}
   \]

   Let's break it down:
   - \(\binom{5}{1} \cdot 4^{12}\) is the number of ways to have at least one bin empty (choose 1 bin to be empty and distribute 12 balls into the remaining 4 bins).
   - \(\binom{5}{2} \cdot 3^{12}\) is the number of ways to have at least two bins empty (choose 2 bins to be empty and distribute 12 balls into the remaining 3 bins).
   - \(\binom{5}{3} \cdot 2^{12}\) is the number of ways to have at least three bins empty (choose 3 bins to be empty and distribute 12 balls into the remaining 2 bins).
   - \(\binom{5}{4} \cdot 1^{12}\) is the number of ways to have at least four bins empty (choose 4 bins to be empty and distribute 12 balls into the remaining 1 bin).

   Plugging in the values:
   \[
   \binom{5}{1} = 5, \quad \binom{5}{2} = 10, \quad \binom{5}{3} = 10, \quad \binom{5}{4} = 5
   \]
   \[
   \left| \bigcup_{i=1}^5 A_i^c \right| = 5^{12} - 5 \cdot 4^{12} + 10 \cdot 3^{12} - 10 \cdot 2^{12} + 5 \cdot 1^{12}
   \]

   Calculating the powers:
   \[
   5^{12} = 244140625, \quad 4^{12} = 16777216, \quad 3^{12} = 531441, \quad 2^{12} = 4096, \quad 1^{12} = 1
   \]

   Substituting back:
   \[
   \left| \bigcup_{i=1}^5 A_i^c \right| = 244140625 - 5 \cdot 16777216 + 10 \cdot 531441 - 10 \cdot 4096 + 5 \cdot 1
   \]
   \[
   = 244140625 - 83886080 + 5314410 - 40960 + 5
   \]
   \[
   = 244140625 - 83886080 + 5314410 - 40960 + 5
   \]
   \[
   = 244140625 - 83886080 + 5273445
assistant