Assuming we have the group: $\{1,2,3,4,5,6\}$ we will create strings where each number will appear once, for example: $123465$ the number $i$ is "in order" if it is in place $i$. In the string above 5,6 are out of order.
1) What is the probability that exactly 3 numbers are out of order?
2) What is the probability that all the numbers in such a string are out of order?
- was (not so easy) the answer is $1/18$
(6 choose 3) ways to put 3 numbers right. Only 2 ways to put the other 3 numbers wrong. that's 10/720
I'm stuck in 2, can this be done without using Inclusion–exclusion principle? It seems too complex.