42. In how many ways can all ten digits be arranged in a row so that the digits $2,5,8$ stand together?
42. We will consider the three digits $2,5,8$ as one number 258. Together with the other seven digits, we will have 8 numbers: $258,0,1,3,4,6,7,9$. There are $8!=40320$ ways to arrange them in a row. In this case, the digits $2,5,8$ will always be together and in the specified order.