Probability numbers from 1-10 I have a problem containing probablility and don't know how to solve it ? 

Numbers from 1 to 10 are ordered randomly.
What is the probability that 1 2 3 appear in the sequence in that particular order - allow other numbers to be between them.

thanks in advance
Hint:

The relevant numbers $1,2,3$ can appear in $3!=6$ different orders that have equal probability to occur.
do not consider 1,2,3 and focus instead on numbers from 4 to 10.
They can occur in 6! ways (ordered randomly).
you have to multiply 6! for the number of ways in which (1,2,3) can present themselves in the sequence( for example, 123 can be at the beginning of the number, in the middle(allow other numbers to be between them)or at the end.
The number of positions of (1,2,3) inside a sequence of ten numbers are the combinations without repetition ten, three.
So favourable cases 6! per C(10,3)/10!(number of possibilities).